x86_64 Linux 中内联 asm 的系统调用？

发布于 2024-07-29 19:37:03 字数 233 浏览 6 评论 0原文

为什么这会打印垃圾而不是优雅地退出我的程序？我在 BSD 上以这种方式使用系统调用，我想知道需要什么才能使其在 Linux 上工作。

int
main(int argc, char **argv)
{
    __asm ("movq $1,%rax; movq $0,%rdi; syscall"); /* exit(0) ? */
    return 0;
}

谢谢。

原文

Why does this print garbage instead of exiting my program gracefully? I use system calls this way on BSD, and I wonder what would I need to make it work in Linux.

int
main(int argc, char **argv)
{
    __asm ("movq $1,%rax; movq $0,%rdi; syscall"); /* exit(0) ? */
    return 0;
}

Thanks.

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葮薆情 2024-08-05 19:37:04

我相信 Syscall 1 在 i386 上退出，但在 x86-64 上写入。

编辑：这似乎不准确：
根据网络，似乎没有太多关于 x86-64 Linux 程序集的信息，这似乎是系统调用指令之前的预期寄存器设置。

 rax  system call number
 rbx  arg0
 rcx  return address from syscall
 rdx  arg2
 rsi  arg3
 rdi  arg4
 r8   arg5
 r9   arg1    (expected by gcc in %rcx)
 r10-r15  should be saved/restored by C code
 rbp  dito What is dito??

Syscall 1 is exit on i386 but write on x86-64 I believe.

EDIT: this seems inaccurate:
According to the web, which does not seem to have too much information about x86-64 Linux assembly this seems to be the expected register setup before the syscall instruction.

 rax  system call number
 rbx  arg0
 rcx  return address from syscall
 rdx  arg2
 rsi  arg3
 rdi  arg4
 r8   arg5
 r9   arg1    (expected by gcc in %rcx)
 r10-r15  should be saved/restored by C code
 rbp  dito What is dito??

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