Linux中是否有任何标准的退出状态代码? [英] Are there any standard exit status codes in Linux?

问题描述

如果某个进程的退出状态为0，则认为该进程已在Linux中正确完成.

A process is considered to have completed correctly in Linux if its exit status was 0.

我已经看到，分段错误通常会导致退出状态为11，尽管我不知道这仅仅是我工作的惯例(失败的应用程序都是内部的)还是标准的.

I've seen that segmentation faults often result in an exit status of 11, though I don't know if this is simply the convention where I work (the apps that failed like that have all been internal) or a standard.

Linux中是否有用于进程的标准退出代码?

Are there standard exit codes for processes in Linux?

推荐答案

在

8 bits of the return code and 8 bits of the number of the killing signal are mixed into a single value on the return from wait(2) & co..

#include <stdio.h>
#include <stdlib.h>
#include <sys/types.h>
#include <sys/wait.h>
#include <unistd.h>
#include <signal.h>

int main() {
    int status;

    pid_t child = fork();
    if (child <= 0)
        exit(42);
    waitpid(child, &status, 0);
    if (WIFEXITED(status))
        printf("first child exited with %u\n", WEXITSTATUS(status));
    /* prints: "first child exited with 42" */

    child = fork();
    if (child <= 0)
        kill(getpid(), SIGSEGV);
    waitpid(child, &status, 0);
    if (WIFSIGNALED(status))
        printf("second child died with %u\n", WTERMSIG(status));
    /* prints: "second child died with 11" */
}

您如何确定退出状态?传统上，shell仅存储8位返回码，但如果进程异常终止，则将高位设置为高位.

How are you determining the exit status? Traditionally, the shell only stores an 8-bit return code, but sets the high bit if the process was abnormally terminated.

$ sh -c 'exit 42'; echo $?
42
$ sh -c 'kill -SEGV $$'; echo $?
Segmentation fault
139
$ expr 139 - 128
11

如果您看到的不是此内容，则程序可能有一个SIGSEGV信号处理程序，该信号处理程序随后会正常调用exit，因此实际上并没有被该信号杀死. (除了SIGKILL和SIGSTOP，程序可以选择处理任何信号.)

If you're seeing anything other than this, then the program probably has a SIGSEGV signal handler which then calls exit normally, so it isn't actually getting killed by the signal. (Programs can chose to handle any signals aside from SIGKILL and SIGSTOP.)

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