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首先發布在這裏。我目前正在開發一個項目，該項目需要編寫一個大型的2d數組（大約1,000,000x7）到我的GPU中，進行一些計算並將其返回給主機。由於我想快速這麼做，並且有這麼大的陣列，所以我試圖壓扁陣列以幫助將其直接傳遞到GPU。數組成功寫入（或者至少cudaMalloc和cudaMemcpy在我寫入設備時返回cudaSuccess），但是當我嘗試讀取它時，cudaMemcpy返回無效的參數錯誤。從設備讀取主機時，cudaMemcpy返回cudaErrorInvalidArgument，不清楚爲什麼
我一直無法弄清楚這是爲什麼，因爲我認爲我應該寫一個有效的1d數組（展平）到設備上並讀出來，並且我認爲我正在將正確的參數提供給做這個。我在網上發現的這個錯誤的唯一結果涉及交換cudaMemcpy的dst和src參數，但我想我已經在這裏了。
這是我的代碼的簡化版本能重現問題：
#include <iostream>
using namespace std;
void alloc2dArray(float ** &arr, unsigned long int rows, unsigned long int cols){
arr = new float*[rows];
arr[0] = new float[rows * cols];
for(unsigned long int i = 1; i < rows; i++) arr[i] = arr[i - 1] + cols;
}
void write2dArrayToGPU(float ** arr, float * devPtr, unsigned long int rows, unsigned long int cols){
if(cudaSuccess != cudaMalloc((void**)&devPtr, sizeof(float) * rows * cols)) cerr << "cudaMalloc Failed";
if(cudaSuccess != cudaMemcpy(devPtr, arr[0], sizeof(float) * rows * cols, cudaMemcpyHostToDevice)) cerr << "cudaMemcpy Write Failed";
}
void read2dArrayFromGPU(float ** arr, float * devPtr, unsigned long int rows, unsigned long int cols){
if(cudaSuccess != cudaMemcpy(arr[0], devPtr, sizeof(float) * rows * cols, cudaMemcpyDeviceToHost)) cerr << "cudaMemcpy Read Failed" << endl;
}
int main(){
int R = 100;
int C = 7;
cout << "Allocating an " << R << "x" << C << " array ...";
float ** arrA;
alloc2dArray(arrA, R, C);
cout << "Assigning some values ...";
for(int i = 0; i < R; i++){
for(int j = 0; j < C; j++){
arrA[i][j] = i*C + j;
}
}
cout << "Done!" << endl;
cout << "Writing to the GPU ...";
float * Darr = 0;
write2dArrayToGPU(arrA, Darr, R, C);
cout << " Done!" << endl;
cout << "Allocating second " << R << "x" << C << " array ...";
float ** arrB;
alloc2dArray(arrB, R, C);
cout << "Done!" << endl;
cout << "Reading from the GPU into the new array ...";
read2dArrayFromGPU(arrB, Darr, R, C);
}
我編譯和我的筆記本電腦與
$nvcc -arch=sm_30 test.cu -o test
$optirun cuda-memcheck ./test
運行此並得到結果：
========= CUDA-MEMCHECK
Allocating an 100x7 array ...Assigning some values ...Done!
Writing to the GPU ... Done!
Allocating second 100x7 array ...Done!
========= Program hit cudaErrorInvalidValue (error 11) due to "invalid argument" on CUDA API call to cudaMemcpy.
========= Saved host backtrace up to driver entry point at error
Reading from the GPU into the new array ...========= Host Frame:/usr/lib64/nvidia-bumblebee/libcuda.so.1 [0x2ef343]
cudaMemcpy Read Failed========= Host Frame:./test [0x38c6f]
========= Host Frame:./test [0x2f08]
========= Host Frame:./test [0x3135]
========= Host Frame:/usr/lib64/libc.so.6 (__libc_start_main + 0xf1) [0x20401]
========= Host Frame:./test [0x2c6a]
=========
========= ERROR SUMMARY: 1 error
我是CUDA的中等新手，仍在學習，所以任何幫助將不勝感激，謝謝！
CUDA是不相關的C. – Olaf
你不能通過按值'devPtr'作爲一個指針參數一個函數，在該指針上執行'cudaMalloc'，然後期望分配的指針值在調用環境中顯示。這是傳遞價值的常見錯誤，當然還有其他類似的問題。比如[this one]（https://stackoverflow.com/questions/22826380/cuda-allocation-and-return-array-from-gpu-to-cpu）。你可能想在那裏研究答案，你的問題可以說是那個答案的重複。 –