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

CUDA - cudaMallocPitch and cudaMemcpy2D use, Error: InvalidValue, InvalidPitchValue

本文关键字：错误 InvalidValue Inval 使用 cudaMallocPitch cudaMemcpy2D CUDA      更新时间：2023-10-16

好的，所以我试图得到一个二维数组cuda的工作，但它成为一个痛苦。错误在标题中，发生在cudaMemcpy2D。我认为这个问题对于训练有素的人来说是显而易见的。提前感谢您的任何帮助，我已经走在了我的班级前面，这是目前的学习指针。

#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;
}

首先，通常您应该为matrix1和matrix2设置单独的间距变量。在这种情况下，它们将是从API调用cudaMallocPitch返回的相同值，但在一般情况下，它们可能不是。

在cudaMemcpy2D行中，呼叫的第二个参数是目标音高。这只是当你做cudaMallocPitch调用这个特定的目标矩阵(即。(第一个参数)

第四个参数是源音高。由于这是用普通的主机分配来分配的，所以除了以字节为单位的宽度之外，它没有间距。

所以你交换了第二个和第四个参数。

所以不用这个:

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);

，对cudaMemcpy2D的第二次调用也是如此。第三次调用实际上是OK的，因为它在相反的方向上，源和目标矩阵被交换，所以它们与你的音调参数正确对齐。

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