问题描述
我希望这个(为了示例而减少)函数可以顺利运行，但由于 fn2 未定义而失败:
I would expect this (reduced for the sake of example) function to run without a hitch, but it fails on account of fn2 is not defined:
void function(){
var var1 = fn1();
var var2 = fn2();
function fn1(){};
return function fn2(){};
}();
return 语句如何从提升中排除 fn2 的函数表达式?
How does the return statement exclude the function expression for fn2 from hoisting?
推荐答案
仅提升使用函数声明创建的函数.return function fn2(){}; 中的函数是使用(命名)函数表达式创建的，因此没有被提升.
Only a function created with a function declaration is hoisted. The function in return function fn2(){}; is created with a (named) function expression so is not hoisted.
函数的评估方式取决于上下文.语句中的任何函数(例如 return 语句)都被解析为函数表达式.另一个例子是在 IIFEs 中使用括号:括号充当分组运算符，确保括号的内容作为表达式求值.
How a function is evaluated is dependent on context. Any function within a statement (such as a return statement) is parsed as a function expression. Another example is the use of parentheses in IIFEs: the parentheses act as a grouping operator, ensuring that the contents of the parentheses are evaluated as an expression.
可以在 Kangax 的优秀文章中找到很多关于此的信息:
Lots of information about this can be found in Kangax's excellent article:
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