Linux，C ++，否定退出代码，说明? [英] Linux,c++, negative exit codes, clarifiction?

问题描述

在Linux中，我们可以返回c ++退出代码吗?即该代码是否有效:

In Linux, c++ can we return exit codes? i.e is this code valid:

int main()
{
  return -1;
}

我要询问是否从bash/zsh编译并执行此代码的原因是，返回值为255，由于使用8位返回码这一事实，所以我将其稀疏，这意味着返回码是只能在正数之间.

The reason I am asking if this code is compiled and executed from bash/zsh, the return value would be 255, I thin this due to the fact that its using an 8 bit return code, and that implies that return codes are only allowed to be between positive.

这正确吗?

推荐答案

相关的系统调用是 _exit()和

The relevant system call is _exit(), and POSIX itself specifies that only the bottom 8 bits of the status given to _exit() are normally usable:

_exit(状态)
状态值可以是 0 ， EXIT_SUCCESS ， EXIT_FAILURE 或任何其他值，尽管只有最低有效8位(即 status& 0377 )应可从 wait()和 waitpid()获得；

_exit(status)
The value of status may be 0, EXIT_SUCCESS, EXIT_FAILURE, or any other value, though only the least significant 8 bits (that is, status & 0377) shall be available from wait() and waitpid();

尽管继续说应该从 waitid()中获得全部值"，但是在Linux上，我似乎仍然只得到低8位.

Though it goes on to say that "the full value shall be available from waitid()", but on Linux, I seem to still get just the low eight bits.

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