用||回显和&& [英] echo with || and &&

问题描述

我在bash上尝试了以下命令

I tried the following command on bash

echo this || echo that && echo other

这给出了输出

this
other

我不明白！

我的空跑是这样的:

1. echo this || echo that && echo other表示true || true && true
2. 由于&&比||具有more precedence，因此第二个表达式优先计算
3. 由于both are true，对||的求值也为true.
4. 因此，我得出的结论是:

1. echo this || echo that && echo other implies true || true && true
2. Since, && has more precedence than ||, the second expression evaluates first
3. Since, both are true, the || is evaluated which also gives true.
4. Hence, I conclude the output to be:

那个

其他

这个

由于Java背景中&&的优先级高于||，所以我无法将其与bash相关.

Being from a Java background where && has more precedence than ||, I am not able to relate this to bash.

任何输入都会很有帮助！

Any inputs would be very helpful!

推荐答案

来自 man bash

3.2.3命令列表

3.2.3 Lists of Commands

列表是由一个或多个流水线组成的序列，这些流水线由运算符';'，'&'，'&&'或'||'中的一个分隔，并可选地以';'中的一个终止，&"或换行符.

A list is a sequence of one or more pipelines separated by one of the operators ‘;’, ‘&’, ‘&&’, or ‘||’, and optionally terminated by one of ‘;’, ‘&’, or a newline.

在这些列表运算符中，&&"和"||"具有相同的优先级，其后是具有相同优先级的;"和&".

Of these list operators, ‘&&’ and ‘||’ have equal precedence, followed by ‘;’ and ‘&’, which have equal precedence.

所以，你的例子

echo this || echo that && echo other

可以读为

(this || that) && other

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