C程序返回codeS和＆功放;＆安培; bash的符号？ [英] C program return codes and && bash symbol?

问题描述

在bash中，我们可以用＆功放;＆安培;运营商执行两个指令。例如：

In bash, we can use the && operator to execute two commands. For example:

./foo && ./bar

将首先执行富，且仅当foo是成功，这将执行栏。但是，当你考虑到C程序，按照惯例，返回这似乎有悖常理0或退出（0）成功完成，运行反直觉的行为时在＆放大器;＆安培; 运营商。 （因为在大多数语言中，0被认为是一个falsey，从而会从停止执行第二条语句。）我缺少什么？

Will first execute foo, and only if foo is "successful", it will then execute bar. However, this seems counter-intuitive when you consider that C programs, by convention, return 0 or exit(0) upon successful completion, which runs counter-intuitive to the behavior of the && operator. (Since in most languages, 0 is considered to be a 'falsey', and thus would stop the second statement from executing.) What am I missing?

推荐答案

您不会错过任何东西。你一定要记住，真正和假不在外壳基本概念。 成功和失败是

You're not missing anything. You just have to keep in mind that true and false aren't fundamental concepts in the shell. success and failure are.

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