Question

该示例代码基于“ cuda”的示例。可以在/NVIDIA_CUDA-9.1_Samples/0_Simple/vectorAdd/vectorAdd.cu

中找到

简而言之，我删除了所有插图，希望不会给您带来麻烦。以下是我的代码。

#include <stdio.h>

__global__ void
vectorAdd( const float *A, const float *B, float *C, int numElements )
{
    int i = blockDim.x * blockIdx.x + threadIdx.x;
    if ( i < numElements )
    {
        C[i] = A[i] + B[i];
    }
    printf( "%d\n", i );
}

int main( void )
{
    int numElements = 50000;
    size_t size = numElements * sizeof( float );
    float *h_A = ( float * ) malloc( size );
    float *h_B = ( float * ) malloc( size );
    float *h_C = ( float * ) malloc( size );
    for ( int i = 0; i < numElements; ++i )
    {
        h_A[i] = rand(  ) / ( float ) RAND_MAX;
        h_B[i] = rand(  ) / ( float ) RAND_MAX;
    }
    float *d_A = NULL;
    cudaMalloc( ( void ** ) &d_A, size );
    float *d_B = NULL;
    cudaMalloc( ( void ** ) &d_B, size );
    float *d_C = NULL;
    cudaMalloc( ( void ** ) &d_C, size );
    cudaMemcpy( d_A, h_A, size, cudaMemcpyHostToDevice );
    cudaMemcpy( d_B, h_B, size, cudaMemcpyHostToDevice );
    int threadsPerBlock = 256;
    int blocksPerGrid = ( numElements + threadsPerBlock - 1 ) / threadsPerBlock;
    printf( "CUDA kernel launch with %d blocks of %d threads\n", blocksPerGrid, threadsPerBlock );
    vectorAdd <<< blocksPerGrid, threadsPerBlock >>> ( d_A, d_B, d_C, numElements );
    cudaMemcpy( h_C, d_C, size, cudaMemcpyDeviceToHost );
    cudaFree( d_A );
    cudaFree( d_B );
    cudaFree( d_C );
    free( h_A );
    free( h_B );
    free( h_C );
    printf( "Done\n" );
    return 0;
}

printf的结果只有4096行，而numElements的数目却没有50000行，并且每次运行都具有不同的范围。

为什么会这样？如何获得正确的printf（50000行，从0〜49999）？感谢您的任何评论和答案。

设备功能中的·printf·功能不会打印每个数字

0 个答案: