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如果我宣佈一個bindless紋理對象與tex1Dfetch和2D紋理
cudaResourceDesc resDesc;
memset(&resDesc, 0, sizeof(resDesc));
resDesc.resType = cudaResourceTypeLinear;
resDesc.res.linear.devPtr = device_global_memory_ptr;
resDesc.res.linear.desc.f = cudaChannelFormatKindUnsigned;
resDesc.res.linear.desc.x = 8 /* 8 bit */ ;
resDesc.res.linear.desc.y = resDesc.res.linear.desc.x;
resDesc.res.linear.desc.z = resDesc.res.linear.desc.x;
resDesc.res.linear.desc.w = resDesc.res.linear.desc.x;
resDesc.res.linear.sizeInBytes = buffer_bytes_size;
cudaTextureDesc texDesc;
memset(&texDesc, 0, sizeof(texDesc));
texDesc.readMode = cudaReadModeElementType;
texDesc.filterMode = cudaFilterModePoint;
texDesc.addressMode[0] = cudaAddressModeBorder;
texDesc.addressMode[1] = cudaAddressModeBorder;
texDesc.addressMode[2] = cudaAddressModeBorder;
cudaTextureObject_t tex1;
cudaCreateTextureObject(&tex1, &resDesc, &texDesc, NULL);
,我以後使用它在CUDA內核
uchar4 pixel = tex1Dfetch<uchar4>(tex1, index);
我仍然會得到一個2D紋理緩存的好處?或者緩存取決於tex1Dfetch
指令?我不能得到上面的代碼與tex2D
不幸地工作。
我很確定答案是否定的。 「2D」緩存需要正確分配,紋理控制器已知寬度/間距。在這種情況下,我認爲你不能這樣做。 – talonmies
@talonmies所以我應該用'cudaMallocArray'分配一個緩衝區,然後爲''cudaMemcpy(device2device)'分配一個緩衝區?我要試試這個,讓你知道 – Dean
@talonmies工作。請讓它成爲答案,我會接受它。 – Dean