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Intro to C FAQ

Contents

Common Errors

General C

Strings

Structs

Types

Arrays

Pointers

Heap, Stack, Dynamic Memory

Questions

runtests.sh: $'\r': command not found or Syntax error: word unexpected (expecting "do")

If you see this error:

Running unit tests:
: not foundtests.sh: 2: ./tests/runtests.sh:
./tests/runtests.sh: 4: ./tests/runtests.sh: Syntax error: word unexpected (expecting "do")

You have two options:

  1. Open the file tests/runtests.sh in VS Code in whatever subproject folder you’re working in, e.g. fizzbuzz. Click on the lower right of the screen where it says CRLF. Choose LF. Save the file. Then the error should go away.

  2. You can do this from the command line with the tr command:

cd tests
cat runtests.sh | tr -d '\r' > runtests.tmp
mv runtests.tmp runtests.sh

The root of the problem is a setting in git that causes all newlines (LF) to be converted to carriage-return/newline (CRLF). The script runtests.sh is a bash script that bash runs, and bash hates \r and pukes everywhere.

To cause git to not do newline conversion for future clones, run the following:

git config --global core.autocrlf false

runtests.sh: 3: 56059 Segmentation fault: 11 $VALGRIND

This means you got a segfault in your program. See What is a segmentation fault and how do I stop it?

Mac: malformed object error when running make tests

This is caused by an older version of the ar and ranlib packages being installed. Sometimes these conflict with the versions installed with xcode.

If running MacPorts:

sudo port selfupdate
sudo port upgrade cctools

If running Brew:

sudo brew update
sudo brew upgrade gcc

Should I bother fixing compiler warnings?

YES!

In C, a warning is the compiler saying, "I can build that, and I will, but it's probably going to do something really messed up that you don't want."

There are only a few warnings you can safely ignore.

If you're absolutely sure the "unused variable" warning is OK, then you could ignore it. Or, better, add a line of code that silences the warning:

void foo(int a)
{
    (void)a; // Do nothing, but at least the compiler will be quiet

Can I accidentally destroy my computer running C code?

Nope! Not with a modern OS.

If you're running MS-DOS, then sure, you can do all kinds of things. I once accidentally blew away all my BIOS settings with a program I wrote and my computer wouldn't boot.

But Windows, Linux, macOS, BSD, or any other mainstream OS from this century all offer memory and resource protection that prevents you from changing memory you're not supposed to, or wiping out a disk you're not supposed to, etc.

The worst you'll see is a Segmentation fault message which means your program tried to do something bad and the OS killed it.

Is a struct comparable to something in Python or JS? Is it like a class?

It's like a class, except with only data (fields, properties) attached to it. There are no methods (functions) associated with it.

If you really want to pretend that you have methods on a struct, you can add them as fields that are pointers to functions. The syntax is pretty obtuse, and it's not a natural or idiomatic thing to do in C.

Example:

struct animal {
    char *name;

    // make_sound is a pointer to a function with no parameters that returns void
    void (*make_sound)(void);
}

// Note how bleat() matches the signature for make_sound(), above
void bleat(void)
{
    printf("Baaaahhhh!\n");
}

int main(void)
{
    struct animal goat;

    // C doesn't have the concept of a constructor, so we have to do it by hand:

    goat.name = "goat";
    goat.make_sound = bleat;

    // Call the "method":

    goat.make_sound(); // Baaaahhhh!
}

Can you have default parameters in the structs?

No. The best you can do is have a helper function set the defaults.

void foo_default(struct foo *f)
{
    f->a = 10; // Set defaults
    f->b = 20;
    f->c = 30;
}
struct foo x;

foo_default(&x); // Set defaults

x.a = 99; // Override default

When you declare a struct, you can also use an initializer to set the field values:

struct foo x = { .a = 10, .b = 20, .c = 30 };

Why does unsigned char type accept a number when it's clearly referring to a character?

Deep down, computers just deal in numbers (1s and 0s). They don't know what a character is. We humans have come up with a system wherein a number represents a certain character. For example, we've agreed that A is 65.

(For information on what number represents what character, look up more detail on ASCII encoding, or its modern superset UTF-8.)

With that in mind, C really only deals in numbers. Even when you put a character in single quotes, it's still just a number. The only difference is in how we interpret that number. That is, is it a value, like 65, or is it a character, like A?

unsigned char c = 'A';

printf("%c\n", c); // Prints "A"
printf("%d\n", c); // Prints 65
unsigned char c = 'A';
int x = c + 10;

printf("%d", x); // Prints 75, since 'A' == 65

In C, whenever you have a character in single quotes like 'A', the compiler treats it just like you'd put the number 65 there. (Or 66 for 'B', and so on.)

The only difference between unsigned char and unsigned int is the number of bytes that are used to represent the number. A char is onGe byte, and an int is typically 4 bytes (but not always).

You can think of these additional bytes as analogous to adding more digits to your numbers. The more digits you have, the more range you can store. Two decimal digits only gets you from 0 to 99, but 8 digits gets you from 0 to 99999999. Similarly, one byte only gets you from 0 to 255, but 4 bytes gets you from 0 to 4,294,967,295.

If you never needed numbers larger than 255, you could use unsigned char for all your variables! (But since modern computers are at least as fast with ints as they are with chars, people just use ints.)

When I pass an array as an argument to a function, when do I use pointer notation and when do I use array notation ?

It's a little-known FunFact that C doesn't actually pass entire arrays to functions. It only passes pointers to the first element in that array.

int a[2000];

// "a" is a pointer to the first element in the array.
// It's the same as &(a[0]).
foo(a);

So when you declare your function, you can do any of these:

void foo(int *a)
void foo(int a[])
void foo(int a[1])
void foo(int a[2000])
void foo(int a[999999999])

and it treats them all as if you'd used:

void foo(int *a)

There's a difference if you want to use multidimensional arrays. You must declare all the dimensions except the first one, which is optional. The compiler needs to know the other dimensions so it can do its array indexing computations correctly.

int foo(int x[][30]) // 30 wide
{
    return x[2][4];
}

int main(void)
{
    int a[10][30]; // 30 wide

    foo(a);

This only applies for multidimensional arrays. For 1-dimensional arrays, the rule still applies; you still need to specify all dimensions except the first one... but since there is only one, you never need to specify it.

Why do functions tend to return pointers to structs, and not just copies of the struct?

It's possible to do this:

struct foo my_func(void)
{
    struct foo f;

    f.x = 10;

    return f; // Return a copy of f
}

as opposed to:

struct foo *my_func(void)
{
    struct foo *p = malloc(sizeof(struct foo));

    p->x = 10;

    return p; // Return a copy of p
}

But in C, it's more idiomatic to return a copy of the pointer to the memory allocated than it is to return a copy of the struct itself.

Part of the reason for this is that it takes time to copy data. A struct can be very large depending on how many fields it has in it, but your average pointer is only 8 bytes.

Since every time you return a thing, a copy of that thing gets made, it is faster to copy a pointer than it is to copy a struct of any non-trivial size.

Finally, note that this variant always invokes undefined behavior and should never be used:

struct foo *my_func(void)
{
    struct foo f;

    f.x = 10;

    return &f; // Return a copy of a pointer to f
}

The reason is because f vaporizes as soon as the function returns (since it's just a local variable), so any pointers to it are invalid.

Why do we subtract '0' from a char to convert it from ASCII to a numeric value?

The code typically looks like this:

char c = '2';  // ASCII '2'

int v = c - '0'; // Convert into numeric value 2

printf("%d\n", v); // prints decimal 2

Remember that in C, a char is like a small int, and when you have a character in single quotes like '2', C replaces that with the ASCII value of that character.

In the case of our example, the ASCII value of '2' is 50. And we want to convert that to the numeric value 2. So we clearly have to subtract 48 from it, since 50 - 48 = 2. But why the '0', then?

Here's part of the ASCII table, just the numbers:

Character ASCII value
'0' 48
'1' 49
'2' 50
'3' 51
'4' 52
'5' 53
'6' 54
'7' 55
'8' 56
'9' 57

It's no coincidence it's done this way. Turns out that if you subtract 48 from any ASCII character that is a digit, you'll end up with the numeric value of that ASCII character.

Example: '7' is value 55 (from the table), compute 55 - 48 and you get 7.

And since '0' is 48, it's become idiomatic in C to convert ASCII digits to values by subtracting '0' from them.

When do I need a pointer to a pointer?

There are a few reasons you might need one, but the most common is when you pass a pointer to a function, and the function needs to modify the pointer.

Let's take a step back and see when you just need to use a pointer.

void foo(int a)
{
    a = 12;
}

int main(void)
{
    int x = 30;

    printf("%d\n", x); // prints 30

    foo(x);

    printf("%d\n", x); // prints 30 again--not 12! Why?
}

In the above example, foo() wants to modify the value of x back in main. But, alas, it can only modify the value of a. When you call a function, all arguments get copied into their respective parameters. a is merely a copy of x, so modifying a has no effect on x.

What if we want to modify x from foo(), though? This is where we have to use a pointer.

void foo(int *a)
{
    *a = 12; // Set the thing `a` points at to 12
}

int main(void)
{
    int x = 30;

    printf("%d\n", x); // prints 30

    foo(&x);

    printf("%d\n", x); // prints 12!
}

In this example, foo() gets a copy of a pointer to x. (Everything gets copied into the parameters when you make a call, even pointers.)

Then it changes the thing the pointer points to to 12. That pointer was pointing to x back in main, so it changes x's value to 12.

Great!

So what about pointers to pointers? It's the same idea. Let's do a broken example:

void alloc_ints(int *p, int count)
{
    p = malloc(sizeof(int) * count); // Allocate space for ints
}

int main(void)
{
    int *q = NULL;

    alloc_ints(q, 10); // Alloc space for 10 ints

    printf("%p\n", q); // Prints NULL still!!

    q[2] = 10;  // UNDEFINED BEHAVIOR, CRASH?
}

What happened?

When we call alloc_ints(), a copy of q is made in p. We then assign into p with the malloc(), but since p is just a copy of q, q is unaffected.

It's just like our first version of foo(), above.

Solution? We need to pass a pointer to q to alloc_ints() so that alloc_ints() can modify the value of q.

But q is already a pointer! It's an int *! So when we take the address-of it (AKA get a pointer to it), we'll end up with a pointer to a pointer, or an int **!

void alloc_ints(int **p, int count)
{
    // Allocate space for ints, store the result in the thing that
    // `p` points to, namely `q`:

    *p = malloc(sizeof(int) * count);
}

int main(void)
{
    int *q = NULL;

    alloc_ints(&q, 10); // Alloc space for 10 ints

    printf("%p\n", q); // Prints some big number, good!

    q[2] = 10;  // works!
}

Success!

Do other languages use pointers?

Most all of them do, but some are more explicit about it than others. In languages like Go, C, C++, and Rust, you have to use the proper operators when using pointers and references.

But languages like JavaScript and Python do a lot of that stuff behind your back. Take this Python example:

class Foo:
    def __init__(self, x):
        self.x = x

def bar(a):
    a.x = 12 # Sets `f.x` to 12--why?

    a = None # Does NOT destroy `f`--why not?


f = Foo(2)

print(f.x) # Prints 2

bar(f)

print(f.x) # Prints 12--why?

Let's look what happened there. We made a new object f, and we passed that object to function bar(), which modified its x property.

After enough time with Python, we learn that it passes objects by reference. This is another way of saying it's using pointers behind your back. Behind the scenes in Python, a is a pointer to f.

That's why when we modify a.x, it actually modifies f.x.

And it's also why when we set a to None, it doesn't change f at all. a is just a pointer to f, not f itself.

Let's look at the C version of that Python program. This works exactly the same way:

#include <stdio.h>

struct foo {
    int x;
};

void bar(struct foo *a)
{
    a->x = 12;   // Sets f.x to 12--why?

    a = NULL;    // Does NOT destroy `f`--why not?
}

int main(void)
{
    struct foo f = { 2 };

    printf("%d\n", f.x); // Prints 2

    bar(&f);

    printf("%d\n", f.x); // Prints 12--why?
}

a is a pointer to f. So we when do a->x, we're saying "set the x property on the thing that a points to".

And when we set a to NULL, it's just modifying a, not the thing that a points to (namely f).

What's the difference between "int *x" and "int* x"?

Syntactically, nothing. They're equivalent.

That said, the recommendation is that you use the form int *x.

Here's why. These two lines are equivalent:

int* x, y;
int *x, y;

In both of them, x is type int*, and y is type int. But by putting the asterisk right next to the int, it makes it look like both x and y are of type int*, when in fact only x is.

If we reverse the order of x and y, we must necessarily move the asterisk with x:

int y, *x; // Also equivalent to the previous two examples

It's idiomatic to keep the asterisk tucked up next to the variable that's the pointer.

What does the "implicit declaration of function" warning mean?

This is the compiler saying "Hey, you're calling a function but I haven't seen a declaration for that function yet." Basically you're calling a function before you've declared it.

If you're calling a library function like printf() or a syscall like stat(), the most common cause of this warning is failure to #include the header file associated with that function. Check the man page for exactly which.

But what if you're getting the error on one of your own functions? Again, it means you're calling that function before you've declared it.

But what does declared mean?

A declaration can either be a function definition, or a function prototype.

Let's look at a broken example:

#include <stdio.h>

int main(void)
{
    foo(); // Implicit declaration warning!!
}

void foo(void)
{
    printf("Foo!\n");
}

In that example, main() calls foo(), but the compiler hasn't seen a declaration of foo() yet. We can fix it by defining foo() before main():

#include <stdio.h>

// Just moved foo()'s definition before main(), that's all

void foo(void)
{
    printf("Foo!\n");
}

int main(void)
{
    foo(); // No problem!
}

You can also use a function prototype to declare a function before it is used, like so:

#include <stdio.h>

void foo(void); // This is the prototype! It's a declaration of foo().

int main(void)
{
    foo(); // No problem
}

void foo(void) // This is the definition of foo()
{
    printf("Foo!\n");
}

Prototypes for functions that are callable from other source files typically go in header files, and then those other source files #include them.

For functions that aren't used outside the current .c file (e.g. little helper functions that no other file will even need to call), those usually are either defined at the top of the file before their first call. If that's inconvenient, a prototype can be placed at the top of the .c file, instead.

What's the difference between puts(), fprintf(), and printf()?

puts() simply outputs a string. It does no formatting of variables. Its only argument is a single string. Additionally, it prints a newline at the end for you.

// This prints "Hello, world %d!" and then a newline:
puts("Hello, world %d!");

printf() does formatted output of variables, and strings as well. It's a superset of puts(), in that way.

int x = 12;

// This prints "Hello, world 12!\n":
printf("Hello, world %d!\n", x);

fprintf() is just like printf(), except it allows you to print to an open file.

FILE *fp = fopen("foo.txt", "w");
int x = 12;

// This writes "Hello, world 12!\n" to the file "foo.txt":
fprintf(fp, "Hello, world %d!\n", x);

Incidentally, there's already a file open for you called stdout (standard output) which normally prints to the screen. These two lines are equivalent:

printf("Hello, world!\n");
fprintf(stdout, "Hello, world!\n"); // Same thing!

There's another already-opened file called stderr (standard error) that is typically used to print error messages. Example:

if (argc != 2) {
    fprintf(stderr, "You must specify a command line argument!\n");
    exit(1);
}

Why does 025 == 21?

In C, any time you have a plain leading 0 on front of a number, the compiler thinks your number is base-8 or octal.

Converting 025 to decimal can be done like so:

2*8 + 5*1 = 16 + 5 = 21

Octal is rarely used in practice, and it's common for new C programmers to put 0 in front of a number in error.

One of the last common places to see octal numbers is in Unix file permissions.

What is the "true dev workflow" in C?

There is none.

Initially, it was in a Unix-like system probably using Makefiles to build the software. This is the system we use at Lambda.

And modern C development under Unix still follows this pattern, except maybe using autotools or CMake.

But dev for specific platforms like Windows probably happens in Visual Studio instead of using make and the rest of it.

Does C have garbage collection?

Nope!

When it comes to freeing up memory that is no longer needed by the program, there are basically two schools of thought:

  • Have the programmer manually manage that memory by explicitly allocating and freeing it. (C's malloc() and free() functions.)
  • Have the runtime automatically manage all that for you. (Garbage collection, automatic reference counting, etc.)

C is too low-level to automatically manage memory usage for you.

One exception is that C automatically allocates and frees local variables just like other languages you're used to. You don't have to explicitly call free() for locals (and it's an error to do so). You must call free for any and all pointers to data that you got back from malloc() when you're done with them.

Also, when a program exits, all memory associated with it is freed by the OS, whether locals or malloc()d data.

Why is C code faster than other languages?

The big thing is interpreted versus compiled.

Python and JavaScript are interpreted languages, which means another program runs your program. It's software running software. So you run python code with the python program and JavaScript code with node, for example.

So in that case, we have the CPU running python, and the Python running your Python program. Python is the middleman, and that takes execution time.

C is a compiled language. The compiler takes your C code, and produces machine code. The CPU runs it directly. No middleman, so it's faster.

But other languages are compiled (like Go, Swift, Rust, C++, and so on). Why is C faster than them, typically?

It's because C is a no-frills, minimalist language. The code you write in C is actually quite close to the machine code that gets produced by the compiler, so it doesn't have to do a lot of things behind your back.

Additionally, people have been working on optimizing the output from C compilers for over 45 years. That's a big head start over other languages.

What is a segmentation fault and how do I stop it?

It means you've accessed some memory you weren't supposed to. The OS killed your process to prevent it from doing so.

The trick is to find the line that's causing the problem. If you get a debugger installed, this can really help.

In lieu of that, use well-positioned printf calls to figure out what the last thing your program does before it crashes.

The bug almost certainly has to do with pointers or arrays (which are just pointers behind syntactic sugar).

Maybe you're accessing a NULL pointer, or an array out of bounds, or modifying something you're not allowed to modify.

What happens if my program exits but I forgot to free() some memory I allocated?

All memory associated with a process is freed when the program exits, even if you forgot to free() it.

It's considered shoddy programming to not free() all the things you malloc()d, though. The OS will free it, but it's bad style to rely on that.

What's the difference between a float and a double, or between an int and a long?

It's all about the range of numbers you want to be able to store.

double can hold a more precise number than a float.

A float might only be precise up to 3.14159, but a double could hold 3.1416925358979, for example.

Likewise, an int might only be able to hold numbers up to 2 billion or so, but a long could hold much larger numbers.

Use as little as you need. If a float or int can do the job, use them. If you need more precision or larger numbers, step up to the next larger type.

Can you use + to concatenate two strings?

No.

The reason is that strings are represented as char* types, and adding two char*s together is not a defined operation in C.

Use the strcat() function in <string.h> to concatenate one string onto another.

Are variables automatically initialized to zero when I declare them?

No.

Always explicitly initialize your variables, whether they be pointers or regular types. If you don't, random garbage will be in them when you use them.

Exception: local variable declared with static storage class (this concept is out of scope for Lambda) and global variables get initialized to zero automatically. But it's still good form to explicitly initialize them.

What type should I use to hold numbers bigger than an int can hold?

If you don't need negative numbers, try unsigned int.

If that's not enough, try long.

If that's not big enough, try long long (yes, that's a real thing).

If those aren't enough, try unsigned long long.

If you just need big numbers, but not a lot of precision, you can use double or long double.

If you need big numbers and a lot of precision and none of the above are big enough, check out the GNU Multiple Precision library. It does arbitrary precision arithmetic to as much precision as you have RAM.

What VS Code plugins are good for C development?

"C/C++ IntelliSense, debugging, and code browsing" by Microsoft is a good one.

What are some additional C resources?

A great C book is The C Programming Language Second Edition, by Kernighan [the "g" is silent] and Ritchie. It's affectionately referred to simply as K&R2.

A less great book that is free online is Beej's Guide to C Programming.

A good, comprehensive FAQ is the comp.lang.c FAQ.

There's no "one true source" of C info online, unfortunately.

Googling printf example, for example, will get you good results.

Googling man printf will bring up the man page for printf.

How do I get the debugger working?

The commonly-used debugger is called gdb (GNU Debugger).

Lambda's own Brian Ruff got it working on the Mac, and made a video covering it.

These instructions are reported good for WSL on Windows.

The CS Wiki page might help, but it's slightly outdated since VS Code is in heavy development.

This video is reported good, as well.

If you're not seeing program output in the Output tab, try adding this to your launch.json:

"externalConsole": true

We recommend Googling for vscode gdb setup macos, substituting whatever platform you're on for macos and setting the search date range to be recent.

How do I print a pointer with printf?

Use the %p format specifier. This will print the value of the pointer (i.e. the memory address), not what it's pointing to (i.e. the value stored at that memory address.)

In practice, pointers are rarely printed except for debugging.

Does C have closures?

No, not in the way that you're used to from higher-level languages. What C does have that essentially acts as a poor man's closure is function pointers. These are literally what their name implies: pointers to functions.

The standard library implementation of the quicksort sorting algorithm provides a good example of this. It's function signature is

void qsort(void *base, size_t nitems, size_t size, int (*compare)(const void *, const void *))

The parameter to notice is the int (*compare)(const void *, const void *). This is a function pointer, mainly denoted by the asterisk in front of the function name all wrapped in parentheses and followed by another set of parentheses containing a parameter signature.

This signature is saying that the qsort function, in addition to the other parameters it receives, also receives a pointer to a function that receives two const void *s and returns an int.

Function pointers are used to fulfill the same use case as closures. They can be used to pass some dynamic logic into another function. In the case of qsort, the function pointer points to a function that specifies how the comparison for the sorting should be done.

If I look at an uninitialized variable, will the garbage in it ever be leftover data from another process?

Not on a modern OS. It would be a security risk, so the OS makes sure this never happens.

How many levels of indirection can you have with pointers? int******?

It's effectively unlimited. But the more you have, the less readable your code is.

In real life:

  • 99.8% (roughly) of pointer usage is single indirection, like int*.
  • 1.5% (roughly) is double indirection, like char**.
  • And the remaining 0.5% is the rest of it.

What's the incompatible integer to pointer conversion error?

This means you have a type mismatch in your assignment.

One side of the = has pointer type, and the other side has integer type.

If you have a pointer in your assignment, both side of the = must be the same pointer type.

Maybe you meant to take the address of the right hand side? Or dereference the right hand side?

Are there any other ways besides malloc() to store things on the heap?

Short answer: no.

(We're assuming that by malloc() we mean malloc(), calloc(), and realloc().)

The longer answer is that you can make a syscall and request more RAM from the operating system. In practice, this is very rare; people just call malloc().

In Unix, that syscall is brk() (or sbrk()). The behavior of this call is a bit strange no

For string literals like "Hello", are those stored on the stack or heap?

Neither.

Consider it to be stored in such a way that it is perpetually accessible from the entire program for the entire run and is never freed. So sort of like the heap.

This code is just fine:

char *hello(void)
{
    char *s = "hello!";

    return s;
}

s is a local variable that is set to point to the string "hello!", and s is deallocated as soon as the function returns. But the data s points to (namely the "hello!") persists for the entire life of the program and is never freed.

It's not actually on the heap, though. The C memory map looks like this, typically:

+--------------------+
|       Stack        |
|         |          |
|         v          |
+- - - - - - - - - - +
|                    |
|                    |
|                    |
+- - - - - - - - - - +
|         ^          |
|         |          |
|        Heap        |
+--------------------+
| Uninitialized data |
+--------------------+
|  Initialized data  |
|    (Read-Write)    |
+--------------------+
|  Initialized data  |
|     (Read-Only)    |
+--------------------+
|    Program code    |
+--------------------+

Constant strings are found in the read-only initialized data section of memory.

If you try to write to one, your program will likely crash:

char *s = "Hello!";

*s = 'B'; // segfault!

Is the C stack like the stack data structure?

Yup! It's used by C to allocate space for local variables when you call functions.

When you return from a function, all those local variables are popped off the stack and thrown away. (Which is why local variables only last as long as the function!)

Is the C heap like a binary heap data structure?

No--it's just a name collision.

Just assume the heap is a big, contiguous chunk of memory. It can be used for whatever, but in C, it is typically managed by malloc() and free() so that we don't have to worry about it.

What are stdin, stdout, and stderr?

These are the three files that are automatically opened for a process when it is first created.

Stream File Name Device
Standard Input stdin Keyboard
Standard Output stdout Screen
Standard Error stderr Screen

stderr is typically used specifically for error messages, even though it goes to the same place as stdout. (The idea is that you can redirect all normal output to one place, and all error output to another place. Or suppress normal output while allowing error output.)

How do I know which header files to #include for any particular function?

Check the man page for the function in question. It'll show it in the Synopsis section.

Example for printf():

SYNOPSIS

   #include <stdio.h>
    int
    printf(const char * restrict format, ...);

Note that if you type man on the command line for a particular function, you might a manual page for another command that isn't the C function. In that case, you have to specify the proper section of the manual for the function.

Try section 3 for library functions, and section 2 for syscalls.

Example looking for printf() in section 3:

man 3 printf

And section 2 for the mkdir() syscall:

man 2 mkdir

When do I have to explicitly cast a type to another type?

Barely ever.

C is pretty good about conversions, and you should be able to build almost everything without casting.

What if you need constant types?

// Print a double:
// (Floating point constants are double by default.)
printf("%lf\n", 3.14);

// Print a float:
printf("%f\n", 3.14f);

// Print a long double
printf("%Lf\n", 3.14L);

// Print a long integer:
printf("%ld\n", 99L);

// Print a long long integer:
printf("%lld\n", 99LL);

// Produce a floating result of a calculation by making sure at least
// one of the operands is a float:
float sort_of_pi = 22.0f / 7;
double double_pi = 22.0 / 7;

What if you need to cast a void pointer?

void foo(void *p)
{
    // convert p to a char*
    char *q = p;

    // Don't need to cast return value from malloc
    int *z = malloc(sizeof(int) * 100);

Some exceptions:

void foo(int a)
{
    // Cast an unused variable to type void to suppress compiler warnings:
    (void)a;

    // If the compiler is warning about an unused return value:
    (void)printf("Hello, world!\n");

    // Cast to a char pointer to iterate over bytes of an object:
    // (C99 6.3.2.3 paragraph 7 allows this.)
    float f = 3.14;
    unsigned char *c = (unsigned char *)&f;

    for (unsigned i = 0; i < sizeof f; i++) {
        printf("%02x ", c[i]);
    }
    printf("\n");

Is realloc() the same as calling malloc(), copying the data over, then calling free() on the original pointer?

Effectively, yes, it's the same. Practically, you should use realloc().

realloc() might be more efficient because in some cases it might not have to copy.

If you grow the space and realloc() knows there's extra unused memory right after the existing space, it will simply tack that addition space onto the end of the memory region and not bother moving the data.

Also, if you shrink the space, realloc() will likely not copy the data. It'll just truncate it.

What happens if I free() a NULL pointer?

Nothing. It's a no-op.

Basically, inside the library code for free(), there's something that looks like this:

void free(void *ptr)
{
    if (ptr == NULL) {
        return;
    }

According to the C99 spec section 7.20.3.2p2:

The free function causes the space pointed to by ptr to be deallocated, that is, made available for further allocation. If ptr is a null pointer, no action occurs. Otherwise, if the argument does not match a pointer earlier returned by the calloc, malloc, or realloc function, or if the space has been deallocated by a call to free or realloc, the behavior is undefined.

What are bits, bytes, kilobytes, megabytes, and all that?

A bit is a single 1 or 0. That's all the numbers it can represent.

A nibble is 4 bits. It can represent numbers from 0b0000 to 0b1111 (binary numbers), which is equivalen to 0 to 15 in decimal.

A byte is 8 bits. It can represent numbers from 0b00000000 to 0b11111111, or 0 to 255 decimal. (Historically, bytes could be other numbers of bits, but on all modern systems, it's always 8 bits. Octet is another term for a number that is specifically 8 bits long.)

A kilobyte is 1024 bytes. (1024 is 210.)

A megabyte is 1024 kilobytes (1,048,576 bytes).

A gigabyte is 1024 megabytes (1,073,741,824 bytes).

A terabyte is 1024 gigabytes (1,099,511,627,776 bytes).

A petabyte is 1024 terabytes (1,125,899,906,842,624 bytes).

If you're used to SI unit prefixes, you might be wondering why in computers kilo means 1024 instead of 1000 like it normally does. In short, it's for historic reasons. 1024 was close enough, so computer programmers adopted the SI prefixes, albeit with a slightly different value.

And that gets confusing. When I say kilobyte, do I mean 1000 bytes or 1024 bytes?

In almost every single case, kilobyte means 1024 bytes. (Hard drive and SSD sizes are sometimes an exception to this rule.)

To remove the ambiguity, you can use a binary prefix, where you'd say kibibyte if you specifically meant 1024 bytes.

That said, in conversation, if someone says kilobyte, odds are extremely high they mean 1024 bytes, not 1000 bytes. kibibyte is uncommonly used in conversation.

In C, can we assume an int is 32 bits?

No.

You can assume an int is at least 16 bits (2 bytes).

There is only one type that has a guaranteed size: sizeof(char) will always be 1 byte. (Same for unsigned char and signed char.)

Never write code that hardcodes or assumes the size of anything other than char. Always use sizeof to get the size.

There's a great Wikipedia article that lists the minimum sizes of the types. If you want your code to be portable to other compilers and systems, choose a type with a minimum size that works for the numbers you need to hold.

What's the difference between #include with double quotes and #include with angle brackets?

In general, use double quotes for your own header files, and angle brackets for built-in system header files like <stdio.h>.

When you #include "foo.h", it looks for foo.h in the same directory as the source file doing the including.

You can also use relative paths, and it'll look relative to the including source file:

#include "../bar.h"
#include "somedir/baz.h"

When you #include <frotz.h>, it looks in the system include directories for the header file. This is where all the built-in header files are installed. On Unix machines, this tends to be the /usr/include directory, but it depends on the OS and compiler.

Should I declare a pointer to a thing, or just declare the thing?

It depends on if you want a thing or not, or if you just want to point to another, already-existing thing.

If there is not an already-existing thing, then making a pointer doesn't make sense. There's no existing thing for it to point to.

This does not declare an int:

int *p;

There's no int there. We have an int pointer, but it's uninitialized, so it points to garbage and can't be used.

So the question to ask is, "Do I already have an existing thing that I can point to? And if so, do I want to point to it?" If the answer to either is "no", then don't use a pointer.

Example:

int a = 12; // here's an existing thing

So the answer to the first part of the question is yes. And do we want a pointer to it? Sure, why not?

int *p = &a; // and there's a pointer to it

Is there a difference between exit() and return?

If you're in the main() function, then no.

If you're in any other function, then yes.

exit() always exits the running process, no matter where you call it from.

If you're in main(), return also exits the running process.

If you're in any other function, return just returns from that function.

Why does strcmp() return 0 when strings match? Since 0 means "false" in C, that seems backwards.

strcmp() returns the difference between two strings. If the strings are the same, there is zero difference, so it returns zero.

This gives strcmp() a little extra power over just returning a boolean true/false value.

For example, if you run this:

strcmp("Antelope", "Buffalo");

it will return less-than zero because "Antelope" is alphabetically less than "Buffalo".

So not only can it tell you if the strings are the same, it can tell you their relative sort order. And that means you can pass it in as the comparator function to the library built-in qsort() function.

What is "undefined behavior" in C?

There are a number of things you're allowed to do in C where the compiler is allowed to produce code that can have any indeterminate effect. It could work, it could crash, it could sort of work, it could crash sometimes and not others, it could crash on some machines and not others.

When you write code that does that, we say the code has undefined behavior.

Wikipedia has a number of practical examples, and if you look in the C99 Language Specification, Annex J.2 you can get a list of all the things you can do that cause undefined behavior.

At Lambda, the most common things you can do to get UB is using bad pointer references.

  • Accessing memory you've already free()d.
  • Freeing memory more than once.
  • Accessing an array off the end of its bounds.
  • Dereferencing a pointer that points to garbage.
  • Dereferencing a NULL pointer.
  • Returning a pointer to a local variable and dereferencing that.

GCC with -Wall -Wextra should warn on a lot of these. This is why it's really important to fix all those warnings.

When you free a pointer, does it get set to NULL automatically?

No.

Furthermore, free() can't do that even if it wanted to.

int *p = malloc(100 * sizeof(int));

free(p);

When we call free(), it gets a copy of the pointer we pass in. (All functions always get copies of all arguments you pass in.) As such, free() could set its copy of p to NULL, but that doesn't affect our original p.

p remains whatever value was in it until we set it to something else.

int *p = malloc(100 * sizeof(int));

free(p);

p = NULL; // NOW p is NULL

(Note that it's undefined behavior to dereference a pointer after you've free()d it. But it's still OK to change that pointer's value.)

How do I write preprocessor macros with #define?

You've probably already seen simple cases of #define like this:

#define antelopes 10

int main(void)
{
    printf("Antelopes: %d\n", antelopes); // prints 10

What's actually happening here is the preprocessor runs through the code before the compiler ever sees it. It manipulates the above code to read:

int main(void)
{
    printf("Antelopes: %d\n", 10); // prints 10

and then hands it off to the compiler. The compiler itself knows nothing about #define.

These #define macros can also accept parameters that make them behave like functions in a way.

Example:

#define square(x) x * x // Not quite Right. See below.

int main(void)
{
    printf("9 squared is %d\n", square(9));

Then the preprocessor generates this code for the compiler:

int main(void)
{
    printf("9 squared is %d\n", 9 * 9);

It just substitutes the parameters in as-is.

Another example:

#define square(x) x * x // Not quite Right. See below.

int main(void)
{
    printf("3 + 2 squared is %d\n", square(3 + 2));

Then the preprocessor generates this code for the compiler, merely substituting in exactly what the dev entered as an argument:

int main(void)
{
    printf("3 + 2 squared is %d\n", 3 + 2 * 3 + 2);

Except that prints 11, when it should print 25 (3 + 2 is 5, and 5 squared is 25)! We have a bug!

Of course, this has to do with the order of operations. We wrote:

3 + 2 * 3 + 2

when what we really wanted was:

(3 + 2) * (3 + 2)

For this reason, you should always put extra parentheses around the macro body, and around every parameter in the body:

#define square(x) ((x) * (x))

And now the expansion of our line will be:

((3 + 2) * (3 + 2))

That will work in all expected cases.

What is an undefined symbol linker error?

This happens when you've called a function, and the linker can't find it in any of the source files or libraries that you're using.

Do you have a warning about an implicit function declaration from the compiler before this error? If so, fix that first.

If not, it could be that you haven't specified all the source files needed on the command line. If you have two sources one.c and two.c, and one calls functions that are in the other, then you have to pass both into the compiler:

gcc -Wall -Wextra -o myexe one.c two.c

Alternately, is there a Makefile present? If so, the author of the software probably intends for you to use that to build the project, rather than trying to figure out the command line on your own.

Try just running:

make

and seeing if that works. Make will show you the command lines it's running to make the build happen so that you don't have to.

The linker is part of the entire compilation system. Basically, the compiler takes your C source files, makes sure they're syntactically correct, and turns them into things called object files, one per source file. These object files might refer to functions that they don't contain, like printf(), for example.

Then the linker takes all the object files and libraries and puts them together into a single binary executable that you can run. It makes sure that all the functions used are present in the files specified.

(Normally this whole process takes place behind the scenes and you don't have to think about it. Sometimes Makefiles will generate object files that you might see, e.g. foo.o. .o is the extension for object files on Unix, or .obj in Windows.)

If the linker can't find a function in any of the object files or libraries, it pukes out an error. It can't call a function if it can't find it.

In this example, the code calls a function foobaz(), but the linker can't find that in any of the object files:

Undefined symbols for architecture x86_64:
  "_foobaz", referenced from:
      _main in foo-133c47.o
ld: symbol(s) not found for architecture x86_64
clang: error: linker command failed with exit code 1 (use -v to see invocation)

(Ignore the leading underscores on the function names.)

To fix, we need to figure out which file foobaz() is defined in, and make sure to pass that filename to the compiler when we build.

How do I make my own header files and what do I put in them?

If you have a .c file and you want to be able to use functions, #defines, or globals from that file in another file, you'll need to make a header file (.h file) to hold the function prototypes, #defines, and references to any globals.

Example:

// square.c

long square(long x)
{
    return x * x;
}

If you want to use square() from a different file, you need a way for that file to know the square() function declaration so the compiler can check to see that you're using it correctly.

For that, we need a corresponding .h file:

// square.h

#ifndef SQUARE_H // Prevent multiple #includes
#define SQUARE_H

// Function prototypes
extern long square(long x);

#endif

The #ifndef SQUARE_H is called the guard macro, and is there to prevent the header file from being included multiple times. (This can be a problem if you have header files that include other header files in a cycle.) #ifndef means "if not defined". Basically, SQUARE_H is acting as a boolean that gets set the first time through and so prevents the content of the header file from being included again.

All header files have wrappers like that. The name of the preprocessor variable is conventionally FILENAME_H, but could be anything as long as it's unique in the project.

The keyword extern indicates to the compiler that the function in question is not defined here; it is defined in another file. In this header file, extern is the default storage class for functions, so it's redundant. But it's really common to see in any case.

Now that we have a header file, we can #include that from somewhere else. Use double quotes in the #include to indicate that the compiler should look in the current directory for the header file.

#include "square.h"

int main(void)
{
    int x = square(5);

When you build, you must specify all .c files on the command line:

gcc -Wall -Wextra -o main main.c square.c

How do I make my own Makefile?

IMPORTANT: Any lines shown indented in any Makefile must begin with a single TAB character! Spaces will not work. If you use spaces to indent, you'll likely see a Missing separator error when you try to make.

myexecutable: mysource1.c mysource2.c
    gcc -Wall -Wextra -o myexecutable mysource1.c mysource2.c

The above Makefile says:

"If mysource1.c or mysource2.c are newer than myexecutable, then run the following commands in the indented block below this line."

And by "newer", we mean they have a more up-to-date timestamp, i.e. they've been saved more recently than myexecutable has been created.

If any of the sources are newer than the executable, the executable should be rebuilt to get it up to date. And that's what make does.

We say that myexecutable depends on mysource1.c and mysource2.c. If either of those dependencies change, myexecutable must be rebuilt.

This also works:

myexecutable: mysource1.c mysource2.c
    gcc -Wall -Wextra -o $@ $^

$@ is a make macro that means "substitute whatever is left of the : right here" (in this example, myexecutable).

$^ is a make macro that means "substitute whatever is to the right of the : right here" (in this example, mysource1.c mysource2.c).

You can also define constants:

SRCS=mysource1.c mysource2.c
TARGET=myexecutable

$(TARGET): $(SRCS)
    gcc -Wall -Wextra -o $@ $^

You can have multiple recipes per Makefile:

target1: source1.c
    gcc -Wall -Wextra -o $@ $^

target2: source2.c
    gcc -Wall -Wextra -o $@ $^

If you type make, it will build the first target in the file by default. A target is a file that is generated by running make, i.e. a file to the left of a :. You can also make specific targets by specifying them on the command line:

$ make target1
gcc -Wall -Wextra -o target1 source1.c
$ make target2
gcc -Wall -Wextra -o target2 source2.c

If you want all targets to get built by default, you can put a dummy first target in that depends on the other targets. This target is called all by convention. It typically doesn't run any commands and is only there to set up the dependency hierarchy with the other recipes.

# Recipe `all` depends on `target1` and `target2`:
all: target1 target2

target1: source1.c
    gcc -Wall -Wextra -o $@ $^

target2: source2.c
    gcc -Wall -Wextra -o $@ $^

.PHONY: all

That .PHONY: all line is a GNU make extension that indicates that all is not a real file. Normally targets refer to real files, and make will check to see if that file exists on disk or not before trying to build it. But in this case, all is not a file; it's just recipe we're using to get all targets to build by default.

Why are there so many printf() variants? How do I know which one to use?

The way to approach it is when you think, "I need something just like printf(), except instead of to the screen, it prints to x," then you look in the man page and see if there's a printf() variant that suits your needs.

The first letters let you know what speciality each one has:

  • printf(): stock, no frills.
  • fprintf(): "file printf"; print to a specified FILE* instead of to stdout.
  • sprintf(): "string printf"; print to a string instead of to stdout.
  • snprintf(): "string printf, with a limited count"; print to a string instead of to stdout, and also specify the maximum number of characters that snprintf() is allowed to store in the buffer. This is good to protect against buffer overruns, and there's a valid argument that you should never use sprintf(), preferring snprintf() instead.
  • vprintf(): "variadic printf"; anything that starts with a v in printf land has to do with variadic functions, i.e. functions with argument lists of variable lengths. These are out of scope at Lambda.
  • etc.

Why is main() always at the bottom of the file?

C has the feature that you have to declare a function before you can use it. So any functions main() needs have to be declared before main(), which means "above" it in the file.

You can also declare functions with prototypes and then put the definition of main() before the definition of the other functions.

It's more common for C devs to put main() at the bottom of the file that contains it, and C devs expect it that way, but it's not wrong or frowned upon to use prototypes to put it at the top instead.

Why does main() return 0? What does the return value mean?

It doesn't have to. The return value from main() is the exit status of the process. This is passed back to the program that first launched your program, so it can do different things based on the exit status of your program.

Think of it like a way for an exiting program to pass a small piece of information (an integer) back to the program that spawned it.

0 by convention means "success". Non-zero means "failure". The idea there is that there's typically only one way for a program to succeed, but many ways for it to fail, and you can communicate those different ways with different exit codes.

Note that you can also use the exit() call, passing an exit status code to it as an argument. Using return in main() is equivalent to calling exit().

In bash, you can look at the exit status of the previous command by echoing the $? shell variable. In the following example, we use grep incorrectly, and it exits with status 2 to indicate that. Whereas we use ls correctly, and it exits with successful status 0:

$ grep
usage: grep [-abcDEFGHhIiJLlmnOoqRSsUVvwxZ] [-A num] [-B num] [-C[num]]
	[-e pattern] [-f file] [--binary-files=value] [--color=when]
	[--context[=num]] [--directories=action] [--label] [--line-buffered]
	[--null] [pattern] [file ...]
$ echo $?
2
$ ls -d .
.
$ echo $?
0

At the OS level, fork() is used to create a new process, and wait() is used to get the exit status back from that process.

Do we have to have a main()? Can there be more than one?

Only if you want to have a program that you can run. When you first launch a program, it looks for a function called main.

There can only be one main() in a program. (There can only be one of any function, for that matter.)

It could be that individual files don't have a main() in them, but when the whole project it built, main() is brought in from another source file.

Also, there's a thing called a library which is a collection of functionality that your program makes use of, but doesn't have a main(), itself. Your program has the main(), and it just calls routines that are in the library.

The C Standard Library is a library that holds all the standard C functionality (e.g. printf(), sqrt(), etc.) but doesn't have a main() of its own. Other programs simply use the library.

Can main() return void? What about main() with no parameters?

No. The function signature for main() must be one of the following:

int main(void)
int main(int argc, char *argv[])
int main(int argc, char **argv)

Note that the second two are equivalent. Use the first form if you don't need to process command line arguments.

Historically, these were also equivalent, but the second form is now obsolete:

int main(void)  // OK
int main()      // Obsolete

Do we need a semicolon at the end of every line?

Yes.

Or, more technically, at the end of every statement or expression.

C won't fill them in automatically like JavaScript will.

Can a pointer point to more than one thing? What about to arrays and structs?

A pointer is a memory address. A single memory address, a single index number into your memory array.

As such, it can only point to a single byte in memory.

For single-byte and multi-byte entities, the pointer always points to the first first byte of that entity.

If you have an int made up of 4 bytes, a pointer to that int points to the address in memory of the first byte of that int.

If you have a struct, it points to the first byte of that struct.

If you have an array of a zillion structs, it points to the first byte of the 0th struct in that array.

If variables are stored in memory, where are pointers stored?

Pointers themselves are variables.

Variables are stored in memory.

Therefore pointers are also stored in memory.

Remember that a pointer is just an index into memory. It's just a number; in fact it's an integer number. We can store integer numbers in memory without a hassle.

The only difference between a pointer and an integer is that you can dereference the pointer to see what it's pointing to. You can't dereference an integer. In memory, they're both just stored as numbers.