如何获取通过 C++ 中的 system() 启动的脚本的退出代码？

Question

我想运行 C++ 应用程序中的脚本并从中捕获退出代码。所以我在应用程序中这样做了：

    std::string script_path = "/home/john/script.sh";
    int i = system(script_path.c_str());
    std::cout << "ERROR: " << i << std::endl;

我写了一个简单的脚本，看它是否能捕捉到错误号：

#!/bin/sh

exit 5

但是程序显示：

ERROR: 1280

我不知道为什么，因为我在脚本中返回了 5。我该如何解决？我用 Linux

Answer 1

How could I fix it? I use Linux

来自man 3 system：

RETURN VALUE

If command is NULL, then a nonzero value if a shell is available, or 0 if no shell is available.

If a child process could not be created, or its status could not be retrieved, the return value is -1 and errno is set to indicate the error.

If a shell could not be executed in the child process, then the return value is as though the child shell terminated by calling _exit(2) with the status 127.

if all system calls succeed, then the return value is the termination status of the child shell used to execute command. (The termination sta‐ tus of a shell is the termination status of the last command it executes.)

In the last two cases, the return value is a "wait status" that can be examined using the macros described in waitpid(2). (i.e., WIFEXITED(), WEXITSTATUS(), and so on).

您可以使用：

std::cout << "ERROR: " << ( (i != -1 && i != 127 && WIFEXITED(i)) ? WEXITSTATUS(i) : -1) << std::endl;

如何获取通过 C++ 中的 system() 启动的脚本的退出代码？

How to get exit code of a script launched via system() in C++?

linux

gcc

g++

sh