Question

我刚开始学习如何使用CUDA。我正在尝试运行一些简单的示例代码：


float *ah, *bh, *ad, *bd;
ah = (float *)malloc(sizeof(float)*4);
bh = (float *)malloc(sizeof(float)*4);
cudaMalloc((void **) &ad, sizeof(float)*4);
cudaMalloc((void **) &bd, sizeof(float)*4);
... initialize ah ...

/* copy array on device */
cudaMemcpy(ad,ah,sizeof(float)*N,cudaMemcpyHostToDevice);
cudaMemcpy(bd,ad,sizeof(float)*N,cudaMemcpyDeviceToDevice);
cudaMemcpy(bh,bd,sizeof(float)*N,cudaMemcpyDeviceToHost);

当我在仿真模式（nvcc -deviceemu）中运行时，它运行正常（并实际复制数组）。但是当我在常规模式下运行时，它会运行w / o错误，但从不复制数据。这就好像忽略了cudaMemcpy行一样。

我做错了什么？

非常感谢，杰森

Answer 1

你应该检查错误，理想情况是在每个malloc和memcpy上，但只是在最后做一次就足够了（cudaGetErrorString(cudaGetLastError())。

只是为了检查显而易见的事情：

你确实有一个支持CUDA的GPU，对吗？运行deviceQuery SDK示例以检查设备是否正常运行，并且所有驱动程序都已安装并正常运行。
N（在memcpy中）等于4（在malloc中），对吧？

Answer 2

查看您是否拥有支持CUDA的设备。您可以尝试运行下面的代码并查看您获得的信息：

#include <cstdio>

int main( void ) {
    cudaDeviceProp  prop;

    int count;
    cudaGetDeviceCount( &count );
    for (int i=0; i< count; i++) {
        cudaGetDeviceProperties( &prop, i );
        printf( "   --- General Information for device %d ---\n", i );
        printf( "Name:  %s\n", prop.name );
        printf( "Compute capability:  %d.%d\n", prop.major, prop.minor );
        printf( "Clock rate:  %d\n", prop.clockRate );
        printf( "Device copy overlap:  " );
        if (prop.deviceOverlap)
            printf( "Enabled\n" );
        else
            printf( "Disabled\n");
        printf( "Kernel execution timeout :  " );
        if (prop.kernelExecTimeoutEnabled)
            printf( "Enabled\n" );
        else
            printf( "Disabled\n" );

        printf( "   --- Memory Information for device %d ---\n", i );
        printf( "Total global mem:  %ld\n", prop.totalGlobalMem );
        printf( "Total constant Mem:  %ld\n", prop.totalConstMem );
        printf( "Max mem pitch:  %ld\n", prop.memPitch );
        printf( "Texture Alignment:  %ld\n", prop.textureAlignment );

        printf( "   --- MP Information for device %d ---\n", i );
        printf( "Multiprocessor count:  %d\n",
                    prop.multiProcessorCount );
        printf( "Shared mem per mp:  %ld\n", prop.sharedMemPerBlock );
        printf( "Registers per mp:  %d\n", prop.regsPerBlock );
        printf( "Threads in warp:  %d\n", prop.warpSize );
        printf( "Max threads per block:  %d\n",
                    prop.maxThreadsPerBlock );
        printf( "Max thread dimensions:  (%d, %d, %d)\n",
                    prop.maxThreadsDim[0], prop.maxThreadsDim[1],
                    prop.maxThreadsDim[2] );
        printf( "Max grid dimensions:  (%d, %d, %d)\n",
                    prop.maxGridSize[0], prop.maxGridSize[1],
                    prop.maxGridSize[2] );
        printf( "\n" );
    }
}

CUDA：cudaMemcpy仅适用于仿真模式

2 个答案: