使用返回javascript函数的new运算符返回奇数范围 [英] Using new operator with return of a javascript function returns odd scope

问题描述

我试图理解为什么新的运行对抗函数而不是示例中的函数返回 y = ：

Im trying to understand why new runs against the function rather than the return of the function in the example y =:

function returnFunction(){ return function blah(str){ this.x = str; return this;}}

y = new returnFunction()("blah")
// output: Window {x: "blah"; top: Window, window: Window, location: Location, ....}
x = new (returnFunction())("blah")
// output: blah {x: "blah"}

z = new function blah(){return this;}()
// output: blah {}

zz = new function(){return this;}() //note the missing function name
// output: Object {}

b = new function blib(str){this.x = str; return this}
// blib {x: undefined}
bb = new function blib(str){this.x = str; return this}("blah")
// blib {x: "blah"}
c = new function blib(){this.x = "blah"; return this}
// blib {x: "blah"}

所以在这种情况下y new 创建 returnFunction 的副本然后调用它

So in the case of y new creates a copy of the returnFunction then invokes it

y = (new returnFunction())()

并且通过调用匿名函数我们没有这个所以它默认为 Window 。

And by invoking an anonymous function we have no this so it defaults to Window.

在 x 的情况下，通过将其包装在parens中（ returnFunction 被调用，返回 blah function）然后 blah 由新的运算符设置调用一个新的对象。

In the case of x, by wrapping it in the parens (returnFunction is invoked returning a the blah function) and then blah is invoked by the new operator setting this to a new object.

我必须包装 new（returnFunction（））来制作它以正确的顺序执行。

It seems odd that I have to wrap new (returnFunction()) to make it execute in the right order.

有人可以向我解释基础执行吗？

Could someone explain to me the underlying execution?

推荐答案

new 的优先级高于调用parens （）。请参阅 https://developer.mozilla.org/en- US / docs / Web / JavaScript / Reference / Operators / Operator_Precedence 。

new has higher precedence than the invocation parens (). See https://developer.mozilla.org/en-US/docs/Web/JavaScript/Reference/Operators/Operator_Precedence.

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