cudaBindTexture2D上的CUDA錯誤

Question

我遇到了cudaBindTexture2D的問題。下面的代碼我創建重現此問題，但這樣的產品......作品：

#include "cuda.h" 
#include <stdio.h> 

// This will output the proper CUDA error strings in the event that a CUDA host call returns an error 
#define checkCudaErrors(err) __checkCudaErrors (err, __FILE__, __LINE__) 

inline void __checkCudaErrors(cudaError err, const char *file, const int line) 
{ 
    if(cudaSuccess != err) 
    { 
     fprintf(stderr, "%s(%i) : CUDA Runtime API error %d: %s.\n",file, line, (int)err, cudaGetErrorString(err)); 
     exit(-1);   
    } 
} 

texture<float,2> myTex; 

int main(int argc, char* argv[]) 
{ 
    float* input; 
    input = new float[656 * 480]; 
    for(int i = 0; i < 656*480; ++i) 
    { 
     input[i] = i; 
    } 
    float* inputDevice; 
    checkCudaErrors(cudaMalloc ((void**)&inputDevice, 656 * 480 * sizeof(float))); 
    checkCudaErrors(cudaMemcpy(inputDevice, input, 656 * 480 * sizeof(float), cudaMemcpyHostToDevice)); 

    cudaChannelFormatDesc desc = cudaCreateChannelDesc<float>(); 

    checkCudaErrors(cudaBindTexture2D(0, myTex, inputDevice, desc, 656, 480, sizeof(float) * 656)); 

    cudaUnbindTexture(myTex); 
    cudaFree(inputDevice); 

    return 0; 
} 

但我真正的項目中，大概是相同的代碼無法正常工作。

texture<float,2> texInput; 

/* a lot of code here, but nothing with texInput */ 

    void CUDAConv::DoConvolution(float* input, float* kernel1D, float* resultMap, unsigned char* rMap, unsigned char* orientMap, int width, int height, int kernelSize) 
    { 
     int fDim = (int)(floor((sqrt((float)(width * width + height * height))/2 + 0.5f))); 
     //Lock* locks = new Lock[width * height]; 
     int dim = fDim * 2 + 1; 
     devWidth = width; 
     devHeight = height; 

     // allocate memory on GPU for the summing images 
     checkCudaErrors(cudaMalloc((void**)&inputDevice, width * height * sizeof(float))); 

     checkCudaErrors(cudaMemcpy(inputDevice, input, width * height * sizeof(float), cudaMemcpyHostToDevice)); 

     cudaChannelFormatDesc desc = cudaCreateChannelDesc<float>(); 
     checkCudaErrors(cudaMalloc((void**)&kernel1DDevice, kernelSize)); 

     checkCudaErrors(cudaBindTexture2D(0, texInput, inputDevice, desc, width, height, sizeof(float) * width)); 

     checkCudaErrors(cudaMalloc((void**)&radiusDevice, width * height)); 
     checkCudaErrors(cudaMalloc((void**)&orientationDevice, width * height)); 
     checkCudaErrors(cudaMalloc((void**)&resultDevice, width * height * sizeof(float))); 
     checkCudaErrors(cudaMemset(resultDevice, 0x00, width * height * sizeof(float))); 
     checkCudaErrors(cudaMemcpy(kernel1DDevice, kernel1D, kernelSize * sizeof(float), cudaMemcpyHostToDevice)); 
     for(int i = 0; i < angles; ++i) 
     { 
      checkCudaErrors(cudaMalloc((void**)&sumUpImage[i], angles * width * height * sizeof(float))); 
      checkCudaErrors(cudaMemset(&sumUpImage[i], 0x00, width * height * angles * sizeof(float))); 
      checkCudaErrors(cudaMalloc((void**)&rotationImage[i], angles * dim * dim * sizeof(float))); 
      checkCudaErrors(cudaMemset(&rotationImage[i], 0x00, dim * dim * angles * sizeof(float))); 
     } 

     // do all convolution calculations in the Convolution function 
     convolution <<<1, angles>>> (/*locks, */inputDevice, kernel1DDevice, rotationImage, sumUpImage, resultDevice, radiusDevice, orientationDevice, devWidth, devHeight, angles); 

     checkCudaErrors(cudaMemcpy(resultMap, resultDevice, width * height * sizeof(float), cudaMemcpyDeviceToHost)); 

     // free memory allocated on the GPU 
     for(int i = 0; i < angles; ++i) 
     { 
      checkCudaErrors(cudaFree(sumUpImage[i])); 
     } 

     //free(locks); 
     cudaUnbindTexture(texInput); 
     cudaFree(inputDevice); 
     cudaFree(kernel1DDevice); 
     cudaFree(radiusDevice); 
     cudaFree(orientationDevice); 
     cudaFree(resultDevice); 
    } 

我會得到結果上cudaBindTexture2D的錯誤是：

CUDA_Conv.cu（203）：CUDA運行時API錯誤18：無效的紋理參考。

當調試texInput看起來就像myTex，我不知道這裏發生了什麼。

在VS2010中使用CUDA4.2。

Answer 1

答案在評論中，我使用了一個我的GPU不兼容的sm，奇怪的是，它不會在cudaMalloc上拋出錯誤，而只會在cudaBindTexture上拋出錯誤，但同時設置相同的sm /他們的表現也一樣。

謝謝羅傑！