JavaScript prototype delegation in function

2018-06-10 03:24:52

This question already has an answer here:

How does JavaScript .prototype work? 21 answers

What you have is a constructor function that has m in its prototype.

var func = function (){};
func.prototype = {m:9};

console.log( func.prototype.m ); // That logs 9

The prototype is assigned not to the function itself but to instances created by this function:

var f = new func();

console.log( f.m ); // That also logs 9

This is where your m is.

For this to also log 9

func.m

you'd have to have m in Function.prototype because the prototype of func is Function

Function.prototype.m = 9;

console.log( func.m ); // Logs 9

By doing func.prototype = {m:9}; , you are setting the 'prototype' property of your func variable to {m:9} .

When you call, func.m , you try to access the 'm' property of your func variable, which you never set before.

Instead, you previously set the 'm' property of the object func.prototype .

If you want to set the property 'm' from your variable func , simply do func.m = 9;

链接地址: http://www.djcxy.com/p/30086.html