x86_64 Linux中的内联asm中的Syscall ?

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.

为什么这个打印垃圾而不是优雅地退出程序?我在BSD使用这种方式调用系统，我想知道我需要什么才能使它在Linux中工作。

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

Thanks.

谢谢。

2 个解决方案

#1

Why does this print garbage instead of exiting my program gracefully?

为什么这个打印垃圾没有优雅地退出我的程序?

Per CESA-2009-001, "Syscall 1 is exit on i386 but write on x86_64".

根据CESA-2009-001，“Syscall 1在i386上退出，但在x86_64上写入”。

what would I need to make it work in Linux

我需要什么使它在Linux中工作

Use the syscall ordinals from the current unistd_64.h

使用来自当前unistd_64.h的syscall序号。

Hope this helps!

希望这可以帮助!

#2

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

Syscall 1在i386上退出，但我相信写在x86-64上。

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.

编辑:这似乎不准确:根据web，它似乎没有太多关于x86-64 Linux程序集的信息，这似乎是syscall指令之前预期的注册设置。

 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??

#1