内存是如何分配堆栈变量？ [英] How is memory allocated for stack variables?

问题描述

在VS（释放），我运行以下命令：

On VS (release), I run the following:

int main(void)
{
   char b[] = "123";
   char a[] = "1234567";
   printf("%x  %x\n", b,a);
   return 0;
}

我可以看到，一对的内存地址为b + 3（字符串的长度）。这表明存储器与没有间隙分配。这保证最少的记忆被使用。
所以，现在我有点相信，所有的编译器将这样做。
我要确保这个猜测这里。有人可以给我一个更正式的证明或告诉我，我的猜测是植根于一个巧合。

I can see that, the mem address of a is b+3(the length of the string). Which shows that the memory are allocated with no gaps. And this guarantee that least memories are used. So, I now kind of believe that all compilers will do so. I want to make sure of this guess here. Can somebody give me an more formal proof or tell me that my guess is rooted on a coincidence.

推荐答案

没有，它不能保证总是会有数据的完美包装。

No, it's not guaranteed that there will always be perfect packing of data.

例如，我编译和拼命地跑在G ++这code，不同的是8。

For example, I compiled and runned this code on g++, and the difference is 8.

您可以阅读更多关于这个 rel=\"nofollow\">。

You can read more about this here.

TL;博士：编译器可以对齐对象在内存中只有某个常数（总是机器字长）地址整除以帮助处理器（对他们来说更容易与这些地址有效）

tl;dr: Compilers can align objects in memory to only addresses divisible by some constant (always machine-word length) to help processor(for them it's easier to work with such addresses)

UPD：大约对准一个有趣的例子：

UPD: one interesting example about alignment:

#include <iostream>
using namespace std;

struct A
{
    int a;
    char b;
    int c;
    char d;
};

struct B
{
    int a;
    int c;
    char b;
    char d;
};

int main()
{
    cout << sizeof(A) << " " << sizeof(B) << "\n";
}

对于我来说，它打印

For me, it prints

16 12

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