的修改到％ESP引起SIGSEGV [英] A modification to %esp cause SIGSEGV

问题描述

有时候我用下面的code，以避免堆栈溢出的编码竞赛参加的时候。

Sometimes I use the following code to avoid stack overflow when taking part in coding competition.

int main()
{
  static const int _STACK_SIZE = MAXN*10;
  static int _STACK[_STACK_SIZE*2], _ESP;
  __asm__ __volatile__
  (
      "movl %%esp, %0\n" 
      "movl %1, %%esp\n":
      "=g"(_ESP):
      "g"(_STACK + _STACK_SIZE):
  );


  // Do Something..


  __asm__ __volatile__
  (  
      "movl %0, %%esp\n":
      :
      "g"(_ESP):
  );
}

据我所知，这ASM code备份％ESP 和移动堆栈 _STACK [] 。

我的问题：为什么这code事业 SIGSEGV 上的x86-64 Linux服务器（它运行良好，对我自己的x86的Linux）？而如何解决？

My Question: Why this code cause SIGSEGV on a x86-64 Linux Server(It runs well on my own x86 Linux)? And how to fix it?

我想，也许是因为％ESP 是一个64位指针??

I guess, maybe it's because %esp is a 64-bit pointer??

我试图删除 __ asm__ __volatile __（MOVL％0，%% ESP \ N：G（_ ESP）：）; ，它似乎运行好？

I tried to delete the __asm__ __volatile__("movl %0, %%esp\n": : "g"(_ESP):); and it seems runs well?

推荐答案

这是因为在x86-64的组装修改32位寄存器，如尤其归零最高32相应的64位寄存器， RSP 在这种情况下，位。在X86-64堆栈指针 RSP 是一个64位寄存器。应该只发生在计算机上的问题，即有超过4吉布 虚拟内存地址空间，因为2 ^ 32字节= 4吉布，所以在计算机有少于4吉布的虚拟内存地址空间，最高的32位将是零呢。

It's because in x86-64 assembly modifying a 32-bit register such as esp zeroes the highest 32 bits of the corresponding 64-bit register, rsp in this case. In x86-64 the stack pointer rsp is a 64-bit register. The problem should occur only in computers that have more than 4 GiB of virtual memory address space, because 2^32 bytes = 4 GiB, so in computers that have less than 4 GiB of virtual memory address space, the highest 32 bits will be zeros anyway.

因此​​，通过修改尤其你让 RSP 指向你的程序没有访问权限的地址，因而导致段错误。在X86-64 code，你平常不使用尤其可言，此时应更换尤其与 RSP 在你的x86-64 code。

So by modifying esp you make rsp point to an address where your program has no access rights, and thus cause a segmentation fault. In x86-64 code you don't normally use esp at all, and you should replace all instances of esp with rsp in your x86-64 code.

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