BLOCK DIMENSION

Accelerated Computing CUDA CUDA Programming and Performance
#1

Hello, :ph34r:

I have a little doubt with the dimension of the block. could this be defined by a variable? I never see that, but I am not sure that it cannot be posible.

dim3 dimbloq (nseq, 1, 1);

where nseq is : int nseq = 32;

I mean, i think i cannot declare dimbloq with variables, i need constant like … ?
[b]
#define DIM_BLOQ 16

dim3 dimbloq (DIM_BLOQ, 1, 1);[/b]

Could this be a real problem?

Thanks a lot!!!

#2

in C:/CUDA/include/vector_types.h

struct dim3

{

unsigned int x, y, z;

#if defined(__cplusplus)

__host__ __device__ dim3(unsigned int x = 1, unsigned int y = 1, unsigned int z = 1) : x(x), y(y), z(z) {}

__host__ __device__ dim3(uint3 v) : x(v.x), y(v.y), z(v.z) {}

__host__ __device__ operator uint3(void) { uint3 t; t.x = x; t.y = y; t.z = z; return t; }

#endif /* __cplusplus */

};

if you use C++ compiler, then

dim3 dimbloq (DIM_BLOQ, 1, 1);

is reasonable, it is constructor.

if you use C compiler, then you can write

dim3 dimbloq ;

dimbloq.x = DIM_BLOQ ;

dimbloq.y = 1 ;

dimbloq.z = 1 ;

#3

Thats a lot.

But you can, for exaple, do something like that…?

int main(){

int n=0;

dim3 dimbloq ;

while (‘some reasonable condition’){

  n++;

}

if (n<32){

dimbloq.x = DIM_BLOQ ;

  dimbloq.y = 1 ;

  dimbloq.z = 1 ;}

call_some_kernel<<<dimgrid, dimbloq1>>> ( … );

}

i think, this can generate troubles, but I am not sure.

Thank you!

Pedro

:ninja:

#4

in fact, you can set

 dimbloq.x =  nx ;

  dimbloq.y = ny ;

  dimbloq.z = nz ;

where nx, ny, nz read from a file

you don’t need to determine value of dimbloq in compilation time, nvcc will

put value of execution configuration into shared memory