c++ CUDA:如何创建2D纹理对象?

nukf8bse  于 9个月前  发布在  其他
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我正在尝试创建2D纹理对象,4x4 uint8_t。下面是代码:

__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;
}

字符串
我预计结果会是这样的:

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,


我做错了什么?

iih3973s

iih3973s1#

当你使用pitch2D变量进行纹理操作时,底层的分配应该是一个适当的 pitched 分配。我认为通常人们会使用cudaMallocPitch来创建它。然而,要求是:
cudaResourceDesc::res::pitch 2D::pitchInline以字节为单位指定两行之间的间距,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,但我打赌你的GPU的属性不是4。但是你在这里指定了4:

resDesc.res.pitch2D.pitchInBytes = 4;

字符串
同样,我认为人们通常会使用使用cudaMallocPitch创建的间距分配来实现这一点。然而,如果行到行的维度(以字节为单位)可以被texturePitchAlignment整除(在我的例子中是32),那么我确实可以传递普通的线性分配。
我做的另一个改变是使用cudaCreateChannelDesc<>()而不是像你一样手动设置参数。这会创建一组不同的desc参数,似乎也会影响结果。研究差异应该不难。
当我调整你的代码来解决这些问题时,我得到的结果对我来说似乎是合理的:

$ 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[ts];
    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
$


正如评论中所问的,如果我要做类似的事情,但使用cudaMallocPitchcudaMemcpy2D,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[ts];
    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,但很容易证明纹理引用也会发生类似的问题。你不能创建任意小的2D纹理引用,就像你不能创建任意小的2D纹理对象一样。我也不打算提供演示,因为它会在很大程度上重复上面的内容,人们不应该在新的开发工作中使用纹理引用-纹理对象是更好的方法。

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