为什么设置-e;真放;＆安培;假放;＆安培;真不退出？ [英] Why does set -e; true && false && true not exit?

问题描述

根据此接受的答案使用设置-e 内建应该足够bash脚本在第一个错误退出。然而，下面的脚本：

According to this accepted answer using the set -e builtin should suffice for a bash script to exit on the first error. Yet, the following script:

#!/usr/bin/env bash

set -e

echo "a"
echo "b"
echo "about to fail" && /bin/false && echo "foo"
echo "c"
echo "d"

打印：

$ ./foo.sh 
a
b
about to fail
c
d

删除回声富 确实停止脚本;但是为什么呢？

removing the echo "foo" does stop the script; but why?

推荐答案

要简化EtanReisner的详细解答，设置-e 仅退出一个'未捕获的错误。你的情况：

To simplify EtanReisner's detailed answer, set -e only exits on an 'uncaught' error. In your case:

echo "about to fail" && /bin/false && echo "foo"

有故障的code， /斌/假，后跟＆放大器;＆安培; 哪些测试它的退出code。由于＆放大器;＆安培; 测试退出code，假定是程序员知道自己在做什么，并预计该命令可能会失败。因此，这个脚本不会退出。

The failing code, /bin/false, is followed by && which tests its exit code. Since && tests the exit code, the assumption is that the programmer knew what he was doing and anticipated that this command might fail. Ergo, the script does not exit.

相比之下，考虑：

echo "about to fail" && /bin/false

该程序不检测或对出口code支线 /斌/假。所以，当 /斌/假失败，设置-e 将导致脚本退出。

The program does not test or branch on the exit code of /bin/false. So, when /bin/false fails, set -e will cause the script to exit.

考虑：

set -e
echo "about to fail" && /bin/false ; echo "foo"

如果 /斌/假失败，此版本将退出。正如情况下＆放大器;＆安培; 使用，最后陈述回声富将因此只能被执行如果 /斌/假是成功的。

This version will exit if /bin/false fails. As in the case where && was used, the final statement echo "foo" would therefore only be executed if /bin/false were to succeed.

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