[clang-tidy] Check request: use std::tie to implement lexicographical comparison

[denzor200]

Assume the structure:

struct A {
    int n;
    std::string s;
    float w;
    float v;
    float d() const noexcept { return w / v; }
};

And this struct has user defined operators with field-by-field manual comparison:

bool operator<(const A& lhs, const A& rhs) noexcept
{
    if (lhs.n != rhs.n) {
        return lhs.n < rhs.n;
    } else if (lhs.s != rhs.s) {
        return lhs.s < rhs.s;
    }
    return lhs.d() < rhs.d();
}

bool operator>(const A& lhs, const A& rhs) noexcept
{
    if (lhs.n != rhs.n) {
        return lhs.n > rhs.n;
    } else if (lhs.s != rhs.s) {
        return lhs.s > rhs.s;
    }
    return lhs.d() > rhs.d();
}

bool operator<=(const A& lhs, const A& rhs) noexcept
{
    return !(lhs > rhs);
}

bool operator>=(const B& lhs, const A& rhs) noexcept
{
    return !(lhs < rhs);
}

We need a check that will suggest to reimplement that operator in a better way using std::tie:

bool operator<(const A& lhs, const A& rhs) noexcept
{
    const auto lhs_d = lhs.d();
    const auto rhs_d = rhs.d();
    return std::tie(lhs.n, lhs.s, lhs_d) < std::tie(rhs.n, rhs.s, rhs_d);
}

bool operator>(const A& lhs, const A& rhs) noexcept
{
    const auto lhs_d = lhs.d();
    const auto rhs_d = rhs.d();
    return std::tie(lhs.n, lhs.s, lhs_d) > std::tie(rhs.n, rhs.s, rhs_d);
}

bool operator<=(const A& lhs, const A& rhs) noexcept
{
    return !(lhs > rhs); // remains unchanged
}

bool operator>=(const B& lhs, const A& rhs) noexcept
{
    return !(lhs < rhs); // remains unchanged
}