CUDA kernel launch macro with templates - c++

Question

I made a macro to simplify CUDA kernel calls:
#define LAUNCH LAUNCH_ASYNC
#define LAUNCH_ASYNC(kernel_name, gridsize, blocksize, ...) \
LOG("Async kernel launch: " #kernel_name); \
kernel_name <<< (gridsize), (blocksize) >>> (__VA_ARGS__);
#define LAUNCH_SYNC(kernel_name, gridsize, blocksize, ...) \
LOG("Sync kernel launch: " #kernel_name); \
kernel_name <<< (gridsize), (blocksize) >>> (__VA_ARGS__); \
cudaDeviceSynchronize(); \
// error check, etc...
Usage:
LAUNCH(my_kernel, 32, 32, param1, param2)
LAUNCH(my_kernel<int>, 32, 32, param1, param2)
This works fine; with the first define I can enable synronous calls and error checking for debugging.
However it does not work with multiple template arguments like below:
LAUNCH(my_kernel<int,float>, 32, 32, param1, param3)
The error message I get in the line where I call the macro:
error : expected a ">"
Is it possible to make this macro work with multiple template arguments?

Answer 1

The problem is that the preprocessor knows nothing about angle bracket nesting, so it interprets the comma between them as macro argument separator.
If the kernel-launch syntax supports parentheses around the kernel name (I can't check now, not on a CUDA machine), you could do this:
LAUNCH((my_kernel<int, float>), 32, 32, param1, param3)

Answer 2

Something else you could try that I have used (based on the macro you posted) is wrapping the kernel block size and grid size arguments in their own macro:
#define KERNEL_ARGS2(grid, block) <<< grid, block >>>
#define KERNEL_ARGS3(grid, block, sh_mem) <<< grid, block, sh_mem >>>
#define KERNEL_ARGS4(grid, block, sh_mem, stream) <<< grid, block, sh_mem, stream >>>
Now you should be able to use your macro like so:
#define CUDA_LAUNCH(kernel_name, gridsize, blocksize, ...) \
kernel_name KERNEL_ARGS2(gridsize, blocksize)(__VA_ARGS__);
You can use it like:
CUDA_LAUNCH(my_kernel, grid_size, block_size, float* input, float* output, int size);
This will launch the kernel called 'my_kernal' with the given grid and block size and the input arguments.

Answer 3

consider this solution that also throws error
inline void echoError(cudaError_t e, const char *strs) {
char a[255];
if (e != cudaSuccess) {
strncpy(a, strs, 255);
fprintf(stderr, "Failed to %s,errorCode %s",
a, cudaGetErrorString(e));
exit(EXIT_FAILURE);
}
}
#define CUDA_KERNEL_DYN(kernel, bpg, tpb, shd, ...){ \
kernel<<<bpg,tpb,shd>>>( __VA_ARGS__ ); \
cudaError_t err = cudaGetLastError(); \
echoError(err, #kernel); \
}