Question

好吧所以我正在努力让一个二维阵列让cuda继续工作，但它变得很痛苦。错误是在标题中，发生在cudaMemcpy2D。我认为这个问题对训练有素的人很明显。提前感谢您的帮助，我已经超越了目前正在学习指针的课程。

#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;
}

Answer 1

首先，一般来说，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cudaMallocPitch的API调用返回的相同值，但在一般情况下它们可能不是。

在cudaMemcpy2D行中，the second parameter to the call是目标音高。这只是您对此特定目标矩阵（即第一个参数）进行cudaMallocPitch调用时返回的音高值。

第四个参数是源音高。由于这是使用普通主机分配进行分配的，因此除了宽度（以字节为单位）外，它没有音高。

所以你交换了第二个和第四个参数。

所以不要这样：

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);

和第二次调用cudaMemcpy2D类似。第三个调用实际上是正常的，因为它是相反的方向，源和目标矩阵是交换的，所以它们正确地与你的音高参数对齐。

CUDA - cudaMallocPitch和cudaMemcpy2D使用，错误：InvalidValue，InvalidPitchValue

1 个答案: