I need help understanding the alignment of data in OpenCL buffers

Question

Given the following structure

typedef struct { float3 position; float8 position1; } MyStruct; 

I create a buffer to pass it as a pointer to the kernel, the buffer will have the previous buffer format.

I understand that I have to add 4 bytes to the buffer after writing three floats to get the next power of two (16 bytes), but I don't understand why I have to add 16 more bytes before writing bytes to position1. Otherwise, I get the wrong values ​​at position1.

Can someone explain to me why?

Answer 1

A float8 is a vector of 8 floats, each float of which is 4 bytes. This is 32 bytes. According to section 6.1.5 of the OpenCL 1.2 specification , type alignment, types are always aligned in size; therefore, float8 must be 32 byte aligned. The same section also states that float3 accepts 4 words. In addition, since sizeof for a struct is designed to allow arrays of structure, it will not shrink from reordering these specific fields. In more complex structures, you can save space by combining smaller fields.