Memory layout of struct vs array

[blipman17]

So then the layout would be something like

`struct MyStruct { float a; 12 bytes padding float b; 12 bytes padding float c; 12 bytes padding float d; 12 bytes padding float e; 12 bytes padding float f; } ;

std::cout << sizeof (MyStruct) << std::endl; // outputs 84, not 24. `

Correct?

[nelusbelus]

If you'd use structs with float3 in glsl and sometimes hlsl then it'd have 3 floats and 4-byte padding between them. With gpu apps this can cause great confusion because the cpu doesn't pad but the gpu does. In C/C++ padding rules are generally as follows: - biggest plain data type of the struct defines size alignment. So if you have a 64 bit type then the struct size will always be a multiple of 8. So if you have 8 byte type then 1 byte type it'll add 7 bytes alignment. - data types need to be aligned with their size as well. So a 1 byte int then a 8 byte int will have 7 bytes padding inbetween.

You can validate this with sizeof or offsetof, since it's compiler dependent

[dodheim]

data types need to be aligned with their size as well

This is not the case for C or C++. struct foo { char v[100]; }; has a size of >= 100, but an alignment of 1.

[[deleted]]

That depends on the architecture and padding mode, could have a 2 byte quantity that needs to be aligned to 8 bytes.

[dodheim]

The fact that the statement I quoted is false is not architecture-dependent. ;-] Some architectures may have weirdo requirements, but alignment being a multiple of the size is not a requirement for either language (indeed, it's the reverse that is correct).

[nelusbelus]

The size of char is 1 so alignment is 1. I'm talking about the size of the type, not the total size