我正在尝试创建2D纹理对象,4x4 uint8_t。这是代码:
__global__ void kernel(cudaTextureObject_t tex)
{
int x = threadIdx.x;
int y = threadIdx.y;
uint8_t val = tex2D<uint8_t>(tex, x, y);
printf("%d, ", val);
return;
}
int main(int argc, char **argv)
{
cudaTextureObject_t tex;
uint8_t dataIn[16] = {0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15};
uint8_t* dataDev = 0;
cudaMalloc((void**)&dataDev, 16);
struct cudaResourceDesc resDesc;
memset(&resDesc, 0, sizeof(resDesc));
resDesc.resType = cudaResourceTypePitch2D;
resDesc.res.pitch2D.devPtr = dataDev;
resDesc.res.pitch2D.desc.x = 8;
resDesc.res.pitch2D.desc.y = 8;
resDesc.res.pitch2D.desc.f = cudaChannelFormatKindUnsigned;
resDesc.res.pitch2D.width = 4;
resDesc.res.pitch2D.height = 4;
resDesc.res.pitch2D.pitchInBytes = 4;
struct cudaTextureDesc texDesc;
memset(&texDesc, 0, sizeof(texDesc));
cudaCreateTextureObject(&tex, &resDesc, &texDesc, NULL);
cudaMemcpy(dataDev, &dataIn[0], 16, cudaMemcpyHostToDevice);
dim3 threads(4, 4);
kernel<<<1, threads>>>(tex);
cudaDeviceSynchronize();
return 0;
}
我希望结果会是这样的:
0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15,
即纹理对象的所有值(顺序无关紧要)。
但实际结果是:
0, 2, 4, 6, 0, 2, 4, 6, 0, 2, 4, 6, 0, 2, 4, 6,
我究竟做错了什么?
当你使用pitch2D
变量进行纹理操作时,底层分配应该是一个合适的音调分配。我认为通常人们会用cudaMallocPitch
创造这个。然而the requirement stated是:
cudaResourceDesc :: res :: pitch2D :: pitchInBytes指定两行之间的间距(以字节为单位),并且必须与cudaDeviceProp :: texturePitchAlignment对齐。
在我的GPU上,最后一个属性是32.我不知道你的GPU,但我敢打赌你的GPU属性不是4。但是你在这里指定4:
resDesc.res.pitch2D.pitchInBytes = 4;
同样,我认为人们通常会使用cudaMallocPitch
创建的分配分配。但是,如果行到行维度(以字节为单位)可以被texturePitchAlignment
(在我的情况下为32)整除,那么我似乎可以传递一个普通的线性分配。
我做的另一个改变是使用cudaCreateChannelDesc<>()
而不是像你那样手动设置参数。这会创建一组不同的desc
参数,并且似乎也会影响结果。研究这些差异应该不难。
当我调整代码来解决这些问题时,我得到的结果对我来说似乎是明智的:
$ cat t30.cu
#include <stdio.h>
#include <stdint.h>
typedef uint8_t mt; // use an integer type
__global__ void kernel(cudaTextureObject_t tex)
{
int x = threadIdx.x;
int y = threadIdx.y;
mt val = tex2D<mt>(tex, x, y);
printf("%d, ", val);
}
int main(int argc, char **argv)
{
cudaDeviceProp prop;
cudaGetDeviceProperties(&prop, 0);
printf("texturePitchAlignment: %lu\n", prop.texturePitchAlignment);
cudaTextureObject_t tex;
const int num_rows = 4;
const int num_cols = prop.texturePitchAlignment*1; // should be able to use a different multiplier here
const int ts = num_cols*num_rows;
const int ds = ts*sizeof(mt);
mt dataIn[ds];
for (int i = 0; i < ts; i++) dataIn[i] = i;
mt* dataDev = 0;
cudaMalloc((void**)&dataDev, ds);
cudaMemcpy(dataDev, dataIn, ds, cudaMemcpyHostToDevice);
struct cudaResourceDesc resDesc;
memset(&resDesc, 0, sizeof(resDesc));
resDesc.resType = cudaResourceTypePitch2D;
resDesc.res.pitch2D.devPtr = dataDev;
resDesc.res.pitch2D.width = num_cols;
resDesc.res.pitch2D.height = num_rows;
resDesc.res.pitch2D.desc = cudaCreateChannelDesc<mt>();
resDesc.res.pitch2D.pitchInBytes = num_cols*sizeof(mt);
struct cudaTextureDesc texDesc;
memset(&texDesc, 0, sizeof(texDesc));
cudaCreateTextureObject(&tex, &resDesc, &texDesc, NULL);
dim3 threads(4, 4);
kernel<<<1, threads>>>(tex);
cudaDeviceSynchronize();
printf("\n");
return 0;
}
$ nvcc -o t30 t30.cu
$ cuda-memcheck ./t30
========= CUDA-MEMCHECK
texturePitchAlignment: 32
0, 1, 2, 3, 32, 33, 34, 35, 64, 65, 66, 67, 96, 97, 98, 99,
========= ERROR SUMMARY: 0 errors
$
正如评论中所要求的那样,如果我要使用cudaMallocPitch
和cudaMemcpy2D
做类似的事情,它可能看起来像这样:
$ cat t1421.cu
#include <stdio.h>
#include <stdint.h>
typedef uint8_t mt; // use an integer type
__global__ void kernel(cudaTextureObject_t tex)
{
int x = threadIdx.x;
int y = threadIdx.y;
mt val = tex2D<mt>(tex, x, y);
printf("%d, ", val);
}
int main(int argc, char **argv)
{
cudaDeviceProp prop;
cudaGetDeviceProperties(&prop, 0);
printf("texturePitchAlignment: %lu\n", prop.texturePitchAlignment);
cudaTextureObject_t tex;
const int num_rows = 4;
const int num_cols = prop.texturePitchAlignment*1; // should be able to use a different multiplier here
const int ts = num_cols*num_rows;
const int ds = ts*sizeof(mt);
mt dataIn[ds];
for (int i = 0; i < ts; i++) dataIn[i] = i;
mt* dataDev = 0;
size_t pitch;
cudaMallocPitch((void**)&dataDev, &pitch, num_cols*sizeof(mt), num_rows);
cudaMemcpy2D(dataDev, pitch, dataIn, num_cols*sizeof(mt), num_cols*sizeof(mt), num_rows, cudaMemcpyHostToDevice);
struct cudaResourceDesc resDesc;
memset(&resDesc, 0, sizeof(resDesc));
resDesc.resType = cudaResourceTypePitch2D;
resDesc.res.pitch2D.devPtr = dataDev;
resDesc.res.pitch2D.width = num_cols;
resDesc.res.pitch2D.height = num_rows;
resDesc.res.pitch2D.desc = cudaCreateChannelDesc<mt>();
resDesc.res.pitch2D.pitchInBytes = pitch;
struct cudaTextureDesc texDesc;
memset(&texDesc, 0, sizeof(texDesc));
cudaCreateTextureObject(&tex, &resDesc, &texDesc, NULL);
dim3 threads(4, 4);
kernel<<<1, threads>>>(tex);
cudaDeviceSynchronize();
printf("\n");
return 0;
}
$ nvcc -o t1421 t1421.cu
$ cuda-memcheck ./t1421
========= CUDA-MEMCHECK
texturePitchAlignment: 32
0, 1, 2, 3, 32, 33, 34, 35, 64, 65, 66, 67, 96, 97, 98, 99,
========= ERROR SUMMARY: 0 errors
$
虽然我们这里有纹理对象,但它很容易证明纹理引用会出现类似的问题。您无法创建任意小的2D纹理参考,就像您无法创建任意小的2D纹理对象一样。我也不会提供相关的演示,因为它会在很大程度上复制上面的内容,并且人们不应该再使用纹理参考来进行新的开发工作 - 纹理对象是更好的方法。