Difference between std::mem_size_of and ::SHADER_SIZE

[davids91]

Anyone would know why is there a difference between std::mem::size_of and ::SHADER_SIZE ( from encase ) ?
For the below struct:

#[derive(Clone, ShaderType)]
pub struct MyData {
    pub a: Vec3,
    pub b: Vec3,
    pub(crate) s: u32,
    pub(crate) dim: u32,
}

this fails:

assert_eq!(
    std::mem::size_of::<MyData>(),
    MyData::SHADER_SIZE.get() as usize
);

with:

assertion `left == right` failed
  left: 32
 right: 48

I tried to calculate the size in wgpu for that struct but I think I messed sth up, because it should be fine...
alignOf(Vec3) should be 16, but upon reordering the struct to:

#[derive(Clone, ShaderType)]
pub struct MyData {
    pub a: Vec3,
    pub(crate) s: u32,
    pub b: Vec3,
    pub(crate) dim: u32,
}

I get the same result

Why might that be and how could one fix it?

[kpreid]

You haven't put #[repr(C)] on your struct, so reordering the fields isn't guaranteed to do anything to its layout.

If you're planning to use encase, why are you worrying about the layout of your Rust structs anyway? The way encase works (if I understand it correctly) is that it copies the fields of your structs into the buffers it manages.

[davids91]

Because I did not know this aspect of encase yet, thank you for the info!

So if I understand correctly it is irrelevant that SHADER_SIZE and std::mem size is different.

[teoxoy]

Yeah, the size difference is to be expected since the layout of WGSL structs is different.

[davids91]

Thank you :)