分配初始化的对齐内存

问题描述

我正在编写一个程序(在C ++中)，在其中我需要分配其起始地址应与缓存行大小对齐的数组.当我分配这些数组时，我还希望将内存初始化为零.

I'm writing a program (in C++) in which I need to allocate arrays whose starting addresses should be aligned with the cache line size. When I allocate these arrays I also want the memory initialized to zero.

现在，我可以使用posix_memalign函数来工作了.这对于获取与内存对齐的数组非常有效，但该数组未初始化.在初始化数组时，是否可以使用更好的功能将数组清零?还是只需要编写一个单独的循环来为我做这些准备?

Right now I have it working using the posix_memalign function. This works well for getting memory aligned arrays but the arrays are uninitilized. Is there a better function I can use to zero out the arrays when I initialize them or do I just have to settle for writing a separate loop to do it for me?

推荐答案

只需在块上调用 memset .确保在调用 memset 之前，不要将指针转换为设置昂贵的类型(例如 char * ).由于您的指针将对齐，因此请确保未从编译器中隐藏信息.

Just call memset on the block. Make sure you don't cast the pointer to a type that's expensive to set (like char *) before calling memset. Since your pointer will be aligned, make sure that information isn't hidden from the compiler.

更新:要澄清我关于不隐藏对齐方式的观点，请比较:

Update: To clarify my point about not hiding alignment, compare:

char* mem_demo_1(char *j)
{ // *BAD* compiler cannot tell pointer alignment, must test
    memset(j, 0, 64);
    return j;
}

char* mem_demo_2(void)
{ // *GOOD* compiler can tell pointer alignment
    char * j = malloc(64);
    memset(j, 0, 64);
    return j;
}

使用 GCC ， mem_demo_1 可以编译为60行汇编，而 mem_demo_2 可以编译为20行.性能差异也很大.

With GCC, mem_demo_1 compiles to 60 lines of assembly while mem_demo_2 compiles to 20. The performance difference is also huge.

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