cuda kernel seems not to be called - cuda

Question

May I know why the following simple cuda program fails on a device with CC5.2?
#include <iostream>
#include <math.h>
#include <stdlib.h>
#define N 1
__global__ void vector_addition( int *a, int *b, int *c )
{
int i = blockDim.x + blockIdx.x + threadIdx.x;
if ( i < N )
c[ i ] = a[ i ] + b[ i ];
}
int main()
{
size_t bytes = N*sizeof( int );
int *A = (int *)malloc(bytes);
int *B = (int *)malloc(bytes);
int *C = (int *)malloc(bytes);
int *d_A, *d_B, *d_C;
cudaMalloc( &d_A, bytes );
cudaMalloc( &d_B, bytes );
cudaMalloc( &d_C, bytes );
for ( int i = 0; i < N; i++ ) {
A[ i ] = 1; B[ i ] = 2; C[ i ] = 0;
}
cudaMemcpy( d_A, A, bytes, cudaMemcpyHostToDevice );
cudaMemcpy( d_B, B, bytes, cudaMemcpyHostToDevice );
int thr_per_blk = 1024;
int blk_in_grid = ceil( float( N ) / thr_per_blk );
vector_addition<<< blk_in_grid, thr_per_blk >>>( d_A, d_B, d_C );
cudaMemcpy( C, d_C, bytes, cudaMemcpyDeviceToHost );
for ( int i = 0; i < N; i++ ) {
if ( C[ i ] != 3 ) {
std::cout << "error\n";
}
}
free( A ); free( B ); free( C );
cudaFree( d_A ); cudaFree( d_B ); cudaFree( d_C );
return 0;
}
The output is error message.

Answer 1

This line in your kernel is not correct:
int i = blockDim.x + blockIdx.x + threadIdx.x;
That is not the proper way to generate a 1D index. It should be:
int i = blockDim.x * blockIdx.x + threadIdx.x;
With your incorrect indexing, the first thread, which should generate a 0 for a globally unique index, generates 1024+0+0 = 1024. This fails the if test in your kernel, so no threads actually do anything.