Does repr(C) apply to the types of fields?

help
1

For example:

struct Foo {
    f1: i8,
    f2: i16,
    f3: i8,
}

#[repr(C)]
struct Bar {
    foo: Foo,
}

in this case, does Foo's layout also be the C representation? If not, How to deal with this problem (in my case the Foo type is not in my crate)?

2

As far as I can tell, no. You can look at the Rust reference.
Could you please tell why you'd need to do this for a type from another crate?

3

The reference only talks about how the fields layout in a struct, not saying (or not clear) how the filed's inner layout. At least the reference is not clear to me.

4

@psionic12, are you just trying to provide a thin, newtype wrapper around some other type from another library? If so, that's what repr(transparent) is for.

5

No, because the declaration of one type shouldn't globally affect the declaration of another type.

4 Likes
6

Rust makes no guarantees on the layout by default. If you need a fixed layout, then Foo would need repr(C).

7

No... I want every thing in the Bar type are represented in C, So I can put them into GPU.

8

This is really painful since I have to create a repr(C) version of data struct for very one in math libraries(vector, quaternion, matrix and so one)... Does anybody who also come across this problem?

9

This sounds like an XY problem. You might want to create a thread about the problem you're trying to solve. You may not need to recreate every struct with repr(C).

10

If the math library you are using doesn't already have #[repr(C)] on its types, consider sending it a suggestion or PR to do so — or switching to one that does. Your requirement is a common one.

1 Like
11

yeah, I found that crate nalgebra makes all the types repr(C), I'm going to use this one instead.

12

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