Cuda 如何在gpu中创建矩阵并在cpu上打印?

Cuda 如何在gpu中创建矩阵并在cpu上打印?,cuda,dynamic-memory-allocation,Cuda,Dynamic Memory Allocation,这是一个在gpu上创建矩阵并在cpu上打印出来的代码。谁能告诉我哪里出了问题。多谢各位 # include <stdio.h> __global__ void create(int **d_a){ int i = threadIdx.x; int j = threadIdx.y; d_a[i][j] = 1; } void errorCheck(){ cudaError_t error = cudaGetLastError(); if

这是一个在gpu上创建矩阵并在cpu上打印出来的代码。谁能告诉我哪里出了问题。多谢各位

# include <stdio.h>


__global__ void create(int **d_a){
    int i = threadIdx.x;
    int j = threadIdx.y;

    d_a[i][j] = 1;
}


void errorCheck(){
    cudaError_t error = cudaGetLastError();
    if(error != cudaSuccess){
        // print the CUDA error message and exit
        printf("CUDA error: %s\n", cudaGetErrorString(error));
        exit(-1);
    }
}

# define N 5
int main(){
    int **d_a, **a;

    a = (int**)malloc(N * sizeof(int*));

    for (int i =0; i < N; i++){
        a[i] = (int*)malloc(N*sizeof(int));
    }

    cudaMalloc((void***)&d_a, N*sizeof(int*));

    for (int i =0; i < N; i++){
        cudaMalloc((void**)&d_a,N*sizeof(int));
    }

    errorCheck();
    create <<<1, N>>>(d_a);
    errorCheck();
    cudaMemcpy(a, d_a, (N*N)*sizeof(int),cudaMemcpyDeviceToHost);


    for (int i =0; i < N; i++ ){
        for (int j = 0; j < N; j++ ){
            printf("%d", a[i][j]);
        }
        printf("\n");
    }

    cudaFree(d_a);
    free(a);

    return 0;
}
#包括
__全局无效创建(int**d\u a){
int i=threadIdx.x;
int j=螺纹内径x.y;
d_a[i][j]=1;
}
无效错误检查(){
cudaError_t error=cudaGetLastError();
如果(错误!=cudaSuccess){
//打印CUDA错误消息并退出
printf(“CUDA错误:%s\n”,cudaGetErrorString(错误));
出口(-1);
}
}
#定义n5
int main(){
国际**d_a,**a;
a=(int**)malloc(N*sizeof(int*);
对于(int i=0;i
内存分配或memcpy是否有问题

内存分配或memcpy是否有问题

两方面都是

这:

cudamaloc((void***)和d_a,N*sizeof(int*);
对于(int i=0;i
必须这样做:

cudaMalloc((void***)&d_a, N*sizeof(int*));
for (int i=0; i < N; i++){
    int *row;
    cudaMalloc((void**)&row, N*sizeof(int));
    cudaMemcpy(d_a+i, &row, sizeof(int*), cudaMemcpyHostToDevice);
}
create <<<1, dim3(N,N)>>>(d_a);
errorCheck();
for(int i=0; i<N; i++) {
    int* row;
    cudaMemcpy(&row, d_a+i, sizeof(int*), cudaMemcpyDeviceToHost);
    cudaMemcpy(a[i], row, sizeof(int) * N, cudaMemcpyDeviceToHost);
}
cudamaloc((void***)和d_a,N*sizeof(int*);
对于(int i=0;i
然后这个:

create <<<1, N>>>(d_a);
errorCheck();
cudaMemcpy(a, d_a, (N*N)*sizeof(int),cudaMemcpyDeviceToHost);
创建(d_a);
错误检查();
cudaMemcpy(a,d_a,(N*N)*sizeof(int),cudaMemcpyDeviceToHost);
必须这样做:

cudaMalloc((void***)&d_a, N*sizeof(int*));
for (int i=0; i < N; i++){
    int *row;
    cudaMalloc((void**)&row, N*sizeof(int));
    cudaMemcpy(d_a+i, &row, sizeof(int*), cudaMemcpyHostToDevice);
}
create <<<1, dim3(N,N)>>>(d_a);
errorCheck();
for(int i=0; i<N; i++) {
    int* row;
    cudaMemcpy(&row, d_a+i, sizeof(int*), cudaMemcpyDeviceToHost);
    cudaMemcpy(a[i], row, sizeof(int) * N, cudaMemcpyDeviceToHost);
}
创建(d_a);
错误检查();
对于(int i=0;i