在CUDA中使用全局对常数内存 | 在CUDA中使用全局对常数内存

本文介绍了在CUDA中使用全局对常数内存的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！

问题描述

在那里，
我有以下代码：

Hey there,I have the following piece of code:

#if USE_CONST == 1
    __constant__ double PNT[ SIZE ];
#else
    __device__ double *PNT;
#endif

稍后我有：

#if USE_CONST == 0
    cudaMalloc((void **)&PNT, sizeof(double)*SIZE);
    cudaMemcpy(PNT, point, sizeof(double)*SIZE, cudaMemcpyHostToDevice);
#else
    cudaMemcpyToSymbol(PNT, point, sizeof(double)*SIZE);
#endif

而点是在代码中定义的某处。当使用 USE_CONST = 1 时，一切都按预期工作，但在没有它工作时，它不工作。我通过

whereas point is somewhere defined in the code before. When working with USE_CONST=1 everything works as expected, but when working without it, than it doesn't. I access the array in my kernel-function via

PNT [index]

这两个变体之间的问题在哪里？
感谢！

Where's the problem between the both variants?Thanks!

推荐答案

CUDA 4.0之前的cudaMemcpyToSymbol的正确用法是：

The correct usage of cudaMemcpyToSymbol prior to CUDA 4.0 is:

cudaMemcpyToSymbol("PNT", point, sizeof(double)*SIZE)

或者：

double *cpnt;
cudaGetSymbolAddress((void **)&cpnt, "PNT");
cudaMemcpy(cpnt, point, sizeof(double)*SIZE, cudaMemcpyHostToDevice);

这可能会更快一点，如果你打算从主机API多次访问符号。

which might be a bit faster if you are planning to access the symbol from the host API more than once.

编辑：误解了这个问题。对于全局内存版本，对常量内存类似于第二个版本。

misunderstood the question. For the global memory version, do something similar to the second version for constant memory

double *gpnt;
cudaGetSymbolAddress((void **)&gpnt, "PNT");
cudaMemcpy(gpnt, point, sizeof(double)*SIZE.  cudaMemcpyHostToDevice););

这篇关于在CUDA中使用全局对常数内存的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持！