//2位数组的长宽分别为height,width
float Src;
int pitch;
//GPU的全局存储器申请了空间src
cudaMallocPitch((void*)&Src, &pitch, (size_t)(width * sizeof(float)), (size_t)height);
//将CPU内存中的data 拷贝到到GPU上的src
//图像已经在float* data中,长宽也为height,width
cudaMemcpy2D(Src,pitch,data,(size_t)width* sizeof(float) ,(size_t)width* sizeof(float), (size_t)height,cudaMemcpyHostToDevice);
//然后绑定到纹理内存2位float的TexInt
//这步有问题吗?
texture<float, 2, cudaReadModeElementType> TexInt;
cudaChannelFormatDesc channelDesc = cudaCreateChannelDesc();
cudaBindTexture2D(0, TexInt, Src, &channelDesc,width * height * sizeof(float),pitch));
//然后访问的时候是用
tex1Dfetch(x,y)
还是tex2D(x,y)?