Render to memory without copy

[elbaro]

I am following Wgpu without a window with the code:

let buffer_slice = output_buffer.slice(..);
let (tx, rx) = futures_intrusive::channel::shared::oneshot_channel();
buffer_slice.map_async(wgpu::MapMode::Read, move |result| {
    tx.send(result).unwrap();
});
self.device.poll(wgpu::Maintain::Wait);
pollster::block_on(rx.receive()).unwrap().unwrap();

let data = buffer_slice.get_mapped_range();
use image::{ImageBuffer, Rgba};
let buffer = ImageBuffer::<Rgba<u8>, _>::from_raw(
    self.size.width,
    self.size.height,
    data,
)
.unwrap();
buffer.save("image.png").unwrap();
output_buffer.unmap();

Now I want to abstract this to fn render(..) -> Vec<u8> but BufferView<'a> is bound on BufferSlice, and on Buffer. Can wgpu directly render to a provided host memory, or can the user take the ownership of the host Buffer data and detach it from wgpu context?

Copying the data from gpu to a host memory (buffer) and then copying again to Vec sounds unnecessary.

[SludgePhD]

Buffer is located in GPU memory, and BufferView is a view into that GPU memory (mapped into your programs address space, so that it can be treated as normal memory). The copy from that to the Vec<u8> would only copy the data to CPU memory once.