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好的，所以我正在尝试为 cuda 制作一个 2D 数组，但这变得很痛苦。错误在标题中并且发生在 cudaMemcpy2D。我认为这个问题对于训练有素的眼睛来说是显而易见的。提前感谢您的帮助，我已经领先于目前正在学习指针的班级。

#include <cuda_runtime.h>
#include <iostream>
#pragma comment (lib, "cudart")

/* Program purpose: pass a 10 x 10 matrix and multiply it by another 10x10 matrix */

float matrix1_host[100][100];
float matrix2_host[100][100];

float* matrix1_device;
float* matrix2_device;  
size_t pitch;
cudaError_t err;

__global__ void addMatrix(float* matrix1_device,float* matrix2_device, size_t pitch){
    // How this works
    // first we start to cycle through the rows by using the thread's ID
    // then we calculate an address from the address of a point in the row, by adding the pitch (size of each row) and  * it by
    // the amount of rows we've already completed, then we can use that address of somewhere at a start of a row to get the colums 
    // in the row with a normal array grab. 

    int r = threadIdx.x;

        float* rowofMat1 = (float*)((char*)matrix1_device + r * pitch);
        float* rowofMat2 = (float*)((char*)matrix2_device + r * pitch);
        for (int c = 0; c < 100; ++c) {
             rowofMat1[c] += rowofMat2[c];
        }

}

void initCuda(){
    err = cudaMallocPitch((void**)matrix1_device, &pitch, 100 * sizeof(float), 100);
    err = cudaMallocPitch((void**)matrix2_device, &pitch, 100 * sizeof(float), 100); 
    //err = cudaMemcpy(matrix1_device, matrix1_host, 100*100*sizeof(float), cudaMemcpyHostToDevice);
    //err = cudaMemcpy(matrix2_device, matrix2_host, 100*100*sizeof(float), cudaMemcpyHostToDevice);
    err = cudaMemcpy2D(matrix1_device, 100*sizeof(float), matrix1_host, pitch, 100*sizeof(float), 100, cudaMemcpyHostToDevice);
    err = cudaMemcpy2D(matrix2_device, 100*sizeof(float), matrix2_host, pitch, 100*sizeof(float), 100, cudaMemcpyHostToDevice);
}

void populateArrays(){
    for(int x = 0; x < 100; x++){
        for(int y = 0; y < 100; y++){
            matrix1_host[x][y] = (float) x + y;
            matrix2_host[y][x] = (float) x + y;
        }
    }
}

void runCuda(){
    dim3 dimBlock ( 100 );
    dim3 dimGrid ( 1 );
    addMatrix<<<dimGrid, dimBlock>>>(matrix1_device, matrix2_device, 100*sizeof(float)); 
    //err = cudaMemcpy(matrix1_host, matrix1_device, 100*100*sizeof(float), cudaMemcpyDeviceToHost);
    err = cudaMemcpy2D(matrix1_host, 100*sizeof(float), matrix1_device, pitch, 100*sizeof(float),100, cudaMemcpyDeviceToHost);
    //cudaMemcpy(matrix1_host, matrix1_device, 100*100*sizeof(float), cudaMemcpyDeviceToHost);
}

void cleanCuda(){
    err = cudaFree(matrix1_device);
    err = cudaFree(matrix2_device);

    err = cudaDeviceReset();
}


int main(){
    populateArrays();
    initCuda();
    runCuda();
    cleanCuda();
    std::cout << cudaGetErrorString(cudaGetLastError());
    system("pause");
    return 0;
}
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1 回答 1

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首先，通常你应该为 matrix1 和 matrix2 设置一个单独的音高变量。在这种情况下，它们将是从 API 调用返回的相同值cudaMallocPitch，但在一般情况下，它们可能不是。

在您的cudaMemcpy2D行中，调用的第二个参数是目标音高。这只是cudaMallocPitch调用此特定目标矩阵（即第一个参数）时返回的音高值。

第四个参数是音源音高。由于这是通过普通主机分配分配的，因此除了以字节为单位的宽度外，它没有间距。

因此，您交换了第二个和第四个参数。

所以而不是这个：

err = cudaMemcpy2D(matrix1_device, 100*sizeof(float), matrix1_host, pitch, 100*sizeof(float), 100, cudaMemcpyHostToDevice);

试试这个：

err = cudaMemcpy2D(matrix1_device, pitch, matrix1_host, 100*sizeof(float), 100*sizeof(float), 100, cudaMemcpyHostToDevice);

同样对于第二次调用cudaMemcpy2D. 第三个调用实际上是可以的，因为它的方向相反，源矩阵和目标矩阵交换了，所以它们与你的音高参数正确对齐。

于 2013-03-15T04:53:42.253 回答