why some code use ptx?

[MonroeD]

for some point, the performance of inline ptx code maybe not better than cuda c code

[jrhemstad]

Could you elaborate on your question? I believe you are asking why does CUB sometimes use inline PTX?

The answer is because sometimes that is the tool we need to get the job done. PTX is a very precise tool that allows us to say exactly what we want to happen and sometimes that's what we need.

[MonroeD]

e.g. this file https://github.com/NVIDIA/cub/blob/e882acc1e709feeb3c96f6f1d8a9fb7af9667823/cub/thread/thread_store.cuh
#define _CUB_STORE_16(cub_modifier, ptx_modifier)
template<>
device forceinline void ThreadStore<cub_modifier, uint4*, uint4>(uint4* ptr, uint4 val)
{
asm volatile ("st."#ptx_modifier".v4.u32 [%0], {%1, %2, %3, %4};" : :
CUB_ASM_PTR(ptr),
"r"(val.x),
"r"(val.y),
"r"(val.z),
"r"(val.w));
}
template<>
device forceinline void ThreadStore<cub_modifier, ulonglong2*, ulonglong2>(ulonglong2* ptr, ulonglong2 val)
{
asm volatile ("st."#ptx_modifier".v2.u64 [%0], {%1, %2};" : :
CUB_ASM_PTR(ptr),
"l"(val.x),
"l"(val.y));
}

can we write follow bellow:

#define _CUB_STORE_16(cub_modifier, ptx_modifier)
template<>
device forceinline void ThreadStore<cub_modifier, uint4*, uint4>(uint4* ptr, uint4 val)
{
ptr[0].x = val.x;
ptr[0].y = val.y;
ptr[0].z = val.z;
ptr[0].x = val.w;
}
template<>
device forceinline void ThreadStore<cub_modifier, ulonglong2*, ulonglong2>(ulonglong2* ptr, ulonglong2 val)
{
ptr[0].x = val.x; \
ptr[0].y = val.y;
}

[jrhemstad]

You could, but you would lose control over the caching behavior of the store operation, see https://docs.nvidia.com/cuda/parallel-thread-execution/index.html#cache-operators

Things like this are what I meant about PTX being a precise tool that gives you more control than CUDA C++ code. These are very low level details that typical user code doesn't need to worry about. Our goal in CUB is to abstract away these details so you don't have to worry about them.