If I do not specify arch and code option, will nvcc automatically decide them?

Accelerated Computing CUDA CUDA NVCC Compiler
1

I wrote a very simple cuda program called vecAdd.cu, the gpu in my machine is L4, when I compile it as follows:

nvcc --verbose vecAdd.cu -o vecAdd

I see that there is a output: “ptxas -arch=sm_52 -m64 “/tmp/tmpxft_00002390_00000000-6_vecAdd.ptx” -o "/tmp/tmpxft_00002390_00000000-10_vecAdd.sm_52.cubin”
But the cc of L4 is actually sm_89, why does nvcc use sm_52 here, is it wrong?

2

SM 5.2 is the default value used by nvcc, see the docs here.

3

Thanks, I see it says:

nvcc x.cu
is equivalent to
nvcc x.cu --gpu-architecture=compute_52 --gpu-code=sm_52,compute_52

Here, my further question is if my machine is L4(sm_89), the built binary code as above will follow JIT execution path which will load ptx code and compile against L4 platform?

4

Yes, although bear in mind that the compiler has optimised based on the sm52 and so may not fully take advantage of features present in sm89 - your driver will JIT compile using sm52 PTX.

If you haven’t already, you may want to check through Section 3.1 of the Programming Guide.