cuda - 我在使用 CUDA 在 GPU 上复制 char 时遇到问题

Question

主持人：

unsigned char exp[128];
unsigned char __e;

i = cudaMalloc( (void**)&__e, 128 * sizeof(unsigned char) );
if(i != cudaSuccess)
    printf("cudaMalloc __e FAIL! Code: %d\n", i);

BN_bn2bin128B(e, exp);  // copy data from e to exp, no problems since here

i = cudaMemcpy( &__e, &exp, 128 * sizeof(unsigned char), cudaMemcpyHostToDevice);
if(i != cudaSuccess)
    printf("cudaMemcpy __e FAIL! Code: %d\n", i);

输出：

cudaMemcpy __e FAIL! Code: 11

错误 11 对应于：

cudaErrorInvalidValue = 11, ///< Invalid value

为什么？错误在哪里？

Answer 1

你声明__e不正确。它必须是一个指针。尝试这个：

unsigned char exp[128];
unsigned char * __e;

i = cudaMalloc( (void**)&__e, 128 * sizeof(unsigned char) );
if(i != cudaSuccess)
    printf("cudaMalloc __e FAIL! Code: %d\n", i);

// whatever goes here to set exp

i = cudaMemcpy( __e, &exp[0], 128 * sizeof(unsigned char), cudaMemcpyHostToDevice);
if(i != cudaSuccess)
    printf("cudaMemcpy __e FAIL! Code: %d\n", i);