Linux，C++，负退出代码，澄清? [英] Linux,c++, negative exit codes, clarifiction?

问题描述

在 Linux 中，c++ 可以返回退出代码吗?即此代码是否有效:

int main(){返回-1；}

我问这个代码是否是从 bash/zsh 编译和执行的原因，返回值将是 255，我认为这是因为它使用 8 位返回码，这意味着返回码是只允许在正数之间.

这对吗?

解决方案

相关的系统调用是_exit()，POSIX 本身指定只有 _exit() 的状态的低 8 位是正常可用的:p><块引用>

_exit(状态)
status 的值可以是 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；

虽然它继续说完整的值应从 waitid() 获得"，但在 Linux 上，我似乎仍然只得到低八位.

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

int main()
{
  return -1;
}

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.

Is this correct?

解决方案

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(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();

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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