它是法律采取的函数参数的地址？ [英] Is it legal to take the address of a function parameter?

问题描述

时的ANSI C明确这code段？
在我的系统（Linux的x86_64的），它似乎运行得很好，并打印地址，但它永远是这样吗？例如。该参数可以通过寄存器传递，并利用该地址看起来不正确。

 的#include＆LT;＆stdio.h中GT;无效美孚（INT一）
{
   的printf（％P \\ N，＆安培; A）;
}INT主要（无效）
{
   富（42）;
   返回0;
}
 

编辑：像GCC外貌将会把由寄存器以它的地址之前传递到堆栈中的值

 ＆LT;富计算值：
  55推RBP
  48 89 E5 MOV RBP，RSP
  48 83 10统分RSP，为0x10
  89 FC 7D MOV DWORD PTR [RBP-位于0x4]，EDI
  B8 1C 06 40 0​​0 MOV EAX，0x40061c
  48 8D 55 FC LEA RDX，[RBP-位于0x4]
  48 89 D6 MOV RSI，RDX
  48 89 C7 MOV偏下，RAX
  B8 00 00 00 00 MOV eax中，为0x0
  E8 D8 FE FF FF通话4003c0＆LT; printf的PLT @＆GT;
  C9休假
  C3 RET
 

解决方案

是的，这是完全合法的 - 当然你不会从该函数返回地址，因为受时间富的回报，这是毫无意义的。

Is this code snippet well defined in ANSI C? On my system (Linux x86_64) it seems to run just fine and print an address, but will it always be the case? E.g. the parameter might be passed via a register, and taking the address of that doesn't seem right.

#include <stdio.h>

void foo(int a)
{
   printf("%p\n", &a);
}

int main(void)
{
   foo(42);
   return 0;
}

EDIT: Looks like GCC will put the value that is passed by register into the stack before taking the address of it.

<foo>:
  55                      push   rbp
  48 89 e5                mov    rbp,rsp
  48 83 ec 10             sub    rsp,0x10
  89 7d fc                mov    DWORD PTR [rbp-0x4],edi
  b8 1c 06 40 00          mov    eax,0x40061c
  48 8d 55 fc             lea    rdx,[rbp-0x4]
  48 89 d6                mov    rsi,rdx
  48 89 c7                mov    rdi,rax
  b8 00 00 00 00          mov    eax,0x0
  e8 d8 fe ff ff          call   4003c0 <printf@plt>
  c9                      leave  
  c3                      ret   

解决方案

Yes, this is perfectly legal - of course you wouldn't return that address from the function, because by the time foo returns, it's meaningless.

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