有没有一个标准，跨入的memcpy的版本？ [英] Is there a standard, strided version of memcpy?

问题描述

我有一个列向量这是10个元素长。我有一个矩阵B是10×10的内存存储B是列为主。我愿与列向量A到覆盖第一个行 B中。

I have a column vector A which is 10 elements long. I have a matrix B which is 10 by 10. The memory storage for B is column major. I would like to overwrite the first row in B with the column vector A.

很显然，我可以这样做：

Clearly, I can do:

for ( int i=0; i < 10; i++ )
{
    B[0 + 10 * i] = A[i];
}

在这里我在离开为零0 + 10 * I 来突出B使用列存储的主要（零是行索引）。

where I've left the zero in 0 + 10 * i to highlight that B uses column-major storage (zero is the row-index).

在今晚CUDA的一些土地有心计，我有一个想法，可能有一个CPU的功能来执行跨进的memcpy？我想在一个低的水平，业绩将取决于是否存在一个跨入加载/存储指令的，我不记得有x86汇编为？

After some shenanigans in CUDA-land tonight, I had a thought that there might be a CPU function to perform a strided memcpy?? I guess at a low-level, performance would depend on the existence of a strided load/store instruction, which I don't recall there being in x86 assembly?

推荐答案

简短的回答：你写的code是一样快，因为它会得到

Short answer: The code you have written is as fast as it's going to get.

龙的回答是：的memcpy 功能是通过一些复杂的内部函数或汇编语言编写，因为它在具有任意大小和对齐内存操作数进行操作。如果你是一个覆盖矩阵的列，那么你的操作数将具有自然对齐，并且您将不再需要求助于同样的招数来获得不俗的速度。

Long answer: The memcpy function is written using some complicated intrinsics or assembly because it operates on memory operands that have arbitrary size and alignment. If you are overwriting a column of a matrix, then your operands will have natural alignment, and you won't need to resort to the same tricks to get decent speed.

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