在CUDA 5.0中,NVIDIA添加了“纹理对象”(
我按照NVIDIA此示例来使用
我原本期望的输出结果是"1,2,3,...,12.",但实际打印出来的是"1,7,7,7,...3,9,..."。
为了验证
cudaTextureObject_t
),使纹理的使用更加容易。以前,必须将纹理定义为全局变量。
我按照NVIDIA此示例来使用
cudaTextureObject_t
。对于1D情况,它可以正常工作。我尝试将示例扩展到适用于2D pitched内存。#define WIDTH 6
#define HEIGHT 2
int width = WIDTH; int height = HEIGHT;
float h_buffer[12] = {1,2,3,4,5,6,7,8,9,10,11,12};
float* d_buffer;
size_t pitch;
cudaMallocPitch(&d_buffer, &pitch, sizeof(float)*width, height);
cudaMemcpy2D(d_buffer, pitch, &h_buffer, sizeof(float)*width, sizeof(float)*width, height, cudaMemcpyHostToDevice);
printf("pitch = %d \n", pitch);
//CUDA 5 texture objects: https://developer.nvidia.com/content/cuda-pro-tip-kepler-texture-objects-improve-performance-and-flexibility
cudaResourceDesc resDesc;
memset(&resDesc, 0, sizeof(resDesc));
resDesc.resType = cudaResourceTypePitch2D;
resDesc.res.pitch2D.devPtr = d_buffer;
resDesc.res.pitch2D.pitchInBytes = pitch;
resDesc.res.pitch2D.width = width;
resDesc.res.pitch2D.height = height;
resDesc.res.pitch2D.desc.f = cudaChannelFormatKindFloat;
resDesc.res.pitch2D.desc.x = 32; // bits per channel
resDesc.res.pitch2D.desc.y = 32;
cudaTextureDesc texDesc;
memset(&texDesc, 0, sizeof(texDesc));
texDesc.readMode = cudaReadModeElementType;
cudaTextureObject_t tex;
cudaCreateTextureObject(&tex, &resDesc, &texDesc, NULL);
为了确定数据是否确实可以通过纹理缓存访问,我在此内核中打印了一些字节:
__global__ void printGpu_tex(cudaTextureObject_t tex) {
int tidx = blockIdx.x * blockDim.x + threadIdx.x;
int tidy = blockIdx.y * blockDim.y + threadIdx.y;
if(tidx < WIDTH && tidy < HEIGHT){
float x = tex2D<float>(tex, tidy, tidx);
printf("tex2D<float>(tex, %d, %d) = %f \n", tidy, tidx, x);
}
}
我原本期望的输出结果是"1,2,3,...,12.",但实际打印出来的是"1,7,7,7,...3,9,..."。
tex2D<float>(tex, 0, 0) = 1.000000
tex2D<float>(tex, 0, 1) = 7.000000
tex2D<float>(tex, 0, 2) = 7.000000
tex2D<float>(tex, 0, 3) = 7.000000
tex2D<float>(tex, 0, 4) = 7.000000
tex2D<float>(tex, 0, 5) = 7.000000
tex2D<float>(tex, 1, 0) = 3.000000
tex2D<float>(tex, 1, 1) = 9.000000
tex2D<float>(tex, 1, 2) = 9.000000
tex2D<float>(tex, 1, 3) = 9.000000
tex2D<float>(tex, 1, 4) = 9.000000
tex2D<float>(tex, 1, 5) = 9.000000
为了验证
d_buffer
数据是否设置正确,我还制作了一个“打印内核”,用于原始的d_buffer
数组,而不使用纹理缓存。__global__ void printGpu_vanilla(float* d_buffer, int pitch) {
int tidx = blockIdx.x * blockDim.x + threadIdx.x;
int tidy = blockIdx.y * blockDim.y + threadIdx.y;
if(tidx < WIDTH && tidy < HEIGHT){
float x = d_buffer[tidy*pitch + tidx];
printf("d_buffer[%d][%d] = %f \n", tidy, tidx, x);
}
}
输出结果看起来很不错(与纹理缓存版本不同):
d_buffer[0][0] = 1.000000
d_buffer[0][2] = 2.000000
d_buffer[0][3] = 3.000000
d_buffer[0][4] = 4.000000
d_buffer[0][5] = 5.000000
d_buffer[0][5] = 6.000000
d_buffer[1][0] = 7.000000
d_buffer[1][6] = 8.000000
d_buffer[1][7] = 9.000000
d_buffer[1][8] = 10.000000
d_buffer[1][9] = 11.000000
d_buffer[1][5] = 12.000000
有什么想法是关于纹理缓存版本出了什么问题吗?
下载链接:
cudaMallocPitch
与cudaMallocArray
之间的区别。在旧的纹理缓存API中,cudaMallocArray
是典型的使用方式。但是,cudaMallocArray
需要一个cudaChannelFormatDesc
,而这在新的cudaTextureObject_t
接口中似乎已经过时了。 - solvingPuzzles