本文介绍了带有退货声明的短路的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!

问题描述

据我了解,逻辑AND&& amp;短路。运算符如下所示:

As far as I understand, short-circuiting with the logical AND && operator works like the following:

假设我有表达式 b 然后 a&& b 相同? b:a ,因为

Assuming I have the expressions a and b then a && b is the same as a ? b : a since

如果 a 是真实的,那么结果将是 b
如果 a 是假的那么结果将是 a (甚至没有尝试解析 b

if a is truthy then the result will be b andif a is falsy then the result will be a (without even trying to resolve b)

这就是为什么以下(演示)代码抛出一个SyntaxError:

That being the case why is the following (demo) code throwing a SyntaxError:

var add = function(a,b) {
  b && return a+b; // if(b) return a+b
  ...
}

有没有办法用return语句短路?

Is there a way to short circuit with a return statement?

推荐答案

&& 二元运算符需要两个部分都是表达式

The && binary operator needs both parts to be expressions.

返回是一个语句而不是一个表达式(它不会产生一个值,因为当函数结束时,一个值就没用了。)

return something is a statement but not an expression (it doesn't produce a value, as a value wouldn't be useful when the function ends).

只需使用

if (b) return a+b;

带有更易于阅读的代码的附加好处。

with the added benefit of an easier to read code.

阅读更多:



  • (EcmaScript规范)

  • (MDN)

  • Expressions vs Statements
  • the return statement (EcmaScript spec)
  • logical operators (MDN)

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08-24 13:54
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