About function pointer on cuda 3.2

I got the code from the forum,it can complie it,but when i run the result is not what we want,does anybody known where is wrong,Thanks!

#include <stdio.h>

typedef float (*op_func_t) (float, float);

device float add_func (float x, float y)
return x + y;

device float mul_func (float x, float y)
return x * y;

// Static pointers to device functions
device op_func_t p_add_func = add_func;
device op_func_t p_mul_func = mul_func;

global void kernel( op_func_t op )
printf(“Result: %f\n”, ( *op )( 1.0, 2.0 ) );

int main()

op_func_t h_add_func;
op_func_t h_mul_func;

// Copy device function pointer to host side
cudaMemcpyFromSymbol( &h_mul_func, p_mul_func, sizeof( op_func_t ) );
cudaMemcpyFromSymbol( &h_add_func, p_add_func, sizeof( op_func_t ) );

op_func_t d_myfunc = h_mul_func;

kernel<<<1,1>>>( d_myfunc );




I compliled the code on my GTX460 && cuda verison 3.2:
nvcc -arch=sm_21 -o 1 test.cu ./1

The result is 0 but What we want is 1.0*2.0 = 2.0

I do not know where is wrong does anyboy hele me,thanks!


You can’t use printf() inside kernel function.

This works only in emulation mode (which is deprecated ;) )

Sorry that is incorrect. Fermi cards (and this is being compiled for a Fermi card) support kernel printf.

to build on what avidday said, cards with compute capability 2.0+ can use printf directly in their kernel function(s), while those with lower compute capabilities can use cuPrintf from the SDK

You’re right. Sorry.

It seems I’m behind the times.