bash - "set -e"不会导致代码块在有条件的情况下退出

Question

我试图弄清楚如果失败的 block 在条件链中，为什么错误保释行为 -e 不会启动:

#!/bin/bash
set -e
{       echo "First"
        ls blat
        echo "(this should not print)"
}

打印出来:

First
ls: cannot access 'blat': No such file or directory

这是正确的。

鉴于以下内容:

#!/bin/bash
set -e
{       echo "First"
        ls blat
        echo "(this should not print)"
} || echo "Encountered an error"

打印出来:

First
ls: cannot access 'blat': No such file or directory
(this should not print)

我希望打印 Encountered an error 而不是 this should not print

有人可以向我解释造成这种差异的原因吗？

Answer 1

来自文档(强调我的):

The shell does not exit if the command that fails is part of the command list immediately following a while or until keyword, part of the test following the if or elif reserved words, part of any command executed in a && or || list except the command following the final && or ||, any command in a pipeline but the last, or if the command's return value is being inverted with !.

因为 ls 是作为 || 列表的非最终部分执行的 {...} 复合的一部分，所以当 ls 具有非零退出状态时，脚本不会退出。