如果在if条件中定义了函数，是否可以将其吊起? [英] Is a function hoisted if it is defined within an if condition?

问题描述

所以假设我有这样的东西

So suppose I have something like this

var x = 1;
   if (function f(){}) {
     x += typeof f;
   }
   x;

这将输出"1undefined".我以为它应该输出"1function"，因为函数f(){}应该已经悬挂在if之上.显然不是这样-为什么?我以为函数声明和主体总是被提升到范围的顶部?

This outputs "1undefined". I thought it should have output "1function", because function f(){} should have been hoisted above the if. This is clearly not the case - why? I thought function declarations and bodies were always hoisted to the top of the scope?

推荐答案

悬挂了函数声明.函数表达式 不是.

Function declarations are hoisted. Function expressions are not.

这将创建一个名为 的函数表达式:

This creates a named function expression:

if(function f(){})

除了检查函数表达式是否为真之外，它什么都不做.(函数表达式总是真实的.)

It doesn't do anything except check to see if the function expression is truthy. (Function expressions are always truthy.)

关于命名函数表达式，请参见 https://kangax.github.io/nfe/#named-expr :

Regarding named function expressions, see https://kangax.github.io/nfe/#named-expr:

要记住的一个重要细节是该名称为，仅在新定义函数的范围

此代码在新函数表达式的范围之外 ，因此 f 是未定义的:

This code is outside the scope of the new function expression, and therefore f is undefined:

x += typeof f;

在命名函数表达式中，您可以毫无问题地引用其名称:

Within a named function expression, you can refer to its name without a problem:

(function f() {
  alert(typeof f);   //function
})();

alert(typeof f);     //undefined

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