Starting out with CUDA, about device code

Question

I am getting started with CUDA programming and I have a question about the kernel coding part. Below is the code I was trying out.
I was trying to get it to print the numbers 1-64 using 8 blocks of 8 threads each. To see that the program is using 8 blocks of 8 threads.

The problem is that my output is something impossibly large and different every time and only one value.

#include <stdio.h>

__global__
void start(int *a){
        *a = blockIdx.x*threadIdx.x*blockDim.x;;
}

int main(){
        int a;
        int *d_a;
        int size = 64*sizeof(int);
        cudaMalloc((void**)&d_a,size);
        cudaMemcpy(d_a,&a,size, cudaMemcpyHostToDevice);
        start<<<8,8>>>(d_a);

        cudaMemcpy(&a,d_a,size,cudaMemcpyDeviceToHost);

        cudaFree(d_a);
        printf("%d\n",a);
        return 0;
}

EDIT: Alright, this is going to sound very dumb, but how do I check if the code was actually sent to the GPU card? I suspect the kernel code isn't being processed at all. Maybe because the GPU is off or something. I am using PUTTY so I don't have physical access to the actual machine.

Answer 1

Two problems, all in the same line of code.

*a = blockIdx.x*threadIdx.x*blockDim.x;;

1. All your threads are writing to the same location. Assuming you want an array containing 1-64 this is not what you want to do. You want something like this:

a[id] = id;

2. Your arithmetic is wrong. If you want your blocks and threads to map into 1-64 you can use this instead

   blockIdx.x*blockDim.x+threadIdx.x;

Putting everything together you can do this:

int id= blockIdx.x*blockDim.x+threadIdx.x;
a[id] = id;