了解Bash短路 [英] Understanding Bash short-circuiting

问题描述

首先，我不是Bash专业人士.几个月前，我发现如果同时使用&&和||短路运算符和花括号，那么如果第一个语句以真实值退出，则如果true块中的最后一个语句以非括号形式退出，零，则失败块也将被执行.像这样:

First of all I'm not a Bash pro. I discovered few months ago that if I use both the && and || short circuit operators in sequence with curly braces, then in case the first statement exits with a truthful value, if the last statement in the true block exits non-zero, then the fail block will be executed too. Like this:

returnNumber 0 && {
    echo 'OK'
    returnNumber 1
} || {
    echo 'NG'
}

将输出:

OK
NG

因此，我为此寻求了最简单的解决方案，并提出了以下建议:

So, I looked for the easiest solution for this, and came up with this:

returnNumber 0 && {
    echo 'OK'
    returnNumber 1
    :
} || {
    echo 'NG'
}

我知道，很容易省略内置的冒号，但这是解决方法的正确方法吗?

I know, it is easy to leave out the colon builtin, but is it a proper way for a workaround?

推荐答案

这实际上是非常常见的重击陷阱.这不是不是错误.

This is actually a very common Bash pitfall. It is not a bug.

returnNumber 0的计算结果为true，因此还要对第二个块(由逻辑和&&连接)进行求值，以确保first && second的结果仍为true.
第二个块输出OK，但计算结果为false，因此现在first && second的结果为false.这意味着还必须对第三部分(用逻辑或||连接)进行评估，从而也显示NG.

returnNumber 0 evaluates to true, so the second block (joined by logical and &&) is evaluated as well to make sure the result of first && second is still true.
The second block outputs OK but evaluates to false, so now the result of first && second is false. This means that the third portion (joined by logical or ||) must be evaluated as well, causing NG to be displayed as well.

您应该使用if语句，而不是依赖&&和||:

Instead of relying on && and ||, you should be using if statements:

if returnNumber 0; then
    echo 'OK'
    returnNumber 1
else
    echo 'NG'
fi

tl; dr:当y可以返回非零退出状态时，切勿使用x && y || z.

tl;dr: Never use x && y || z when y can return a non-zero exit status.

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