在AMD64上简单分配两个缓冲区

问题描述

- 由GCC编制的代码：


int main（无效）

{

char a [] =" 123" ;;

char b [] =" abc";


printf（"％p％p \ n"，a，b） ;


返回0;

}


- 在AMD64上运行它给了我：


0x7fff929c9410 0x7fff929c9400

- 两个地址和

之间的16字节距离不仅仅是8个字节的原因是什么？

解决方案

询问gcc实现者。是他们决定编译器应该如何分配空间，而不是C语言本身。

C不需要16字节分隔，但不是

禁止它。


-

Eric Sosman
lid

将'a''和'b''置为无效*。

这是标准C未提及的实现细节。由于对齐原因，它可能是b $ b。

我知道。只是对8字节16字节的选择的好奇心（我希望b $ b预计）。


无论如何，谢谢。

- Code compiled by GCC:

int main(void)
{
char a[] = "123";
char b[] = "abc";

printf("%p %p\n", a, b);

return 0;
}

- Running it on AMD64 gives me:

0x7fff929c9410 0x7fff929c9400

- What is the reason for 16 bytes of distance between two address and
not just 8 bytes?

解决方案

Ask the gcc implementors. It was they who decided how
the compiler should allocate space, not the C language itself.
C does not require a sixteen-byte separation, but does not
forbid it.

--
Eric Sosman
es*****@ieee-dot-org.invalid

Cast ''a'' and ''b'' to void *.

It''s an implementation detail that Standard C says nothing about. It may
be due to alignment reasons.

I know. Just curiosity about the election of 16 bytes over 8 (what I
expected).

Thanks anyway.

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