为什么在shell中0为真而假为1? [英] Why 0 is true but false is 1 in the shell?

问题描述

false; echo $?

上面会输出1，这和我知道的所有其他编程语言都是矛盾的.

The above will output 1, which is contradictory with all other programming languages I know.

这有什么原因吗?

推荐答案

这是一种约定，但仔细想想，它特别有用.一般来说，如果一个程序成功了，这就是你需要知道的一切.但是，如果失败，您可能需要了解有关失败的各种信息——为什么会发生、如何修复等.零均值成功"和非零均值失败让您可以很容易地检查是否成功，并根据需要调查特定错误以获取更多详细信息.许多 API 和框架都有类似的约定 - 成功的函数返回 0，失败的函数返回描述特定失败情况的错误代码.

It's a convention, but a particularly useful one when you think about it. In general, if a program succeeds that's all you need to know. If it fails, however, you might need to know all kinds of information about the failure - why it happened, how to fix it, etc. Having zero mean 'success' and non-zero mean failure lets you can check pretty easily for success, and investigate the particular error for more details if you want to. A lot of APIs and frameworks have a similar convention - functions that succeed return 0 and and those that fail give back an error code describing the particular failure case.

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