为什么对象中的函数返回未定义？

Question

Consider this code:

var MSE = {
    Module : {}
};

MSE.Module = (function() {

    'use-strict';

    var app = {

        tabsPre : function() {

            var tabsPre = {
                init : function() {

                },
                changeTab : function(arg) {
                    return arg;
                }
            };

            tabsPre.init();
            return tabsPre;
        }
    };

    return app;

})();

console.log( MSE.Module.tabsPre() );
console.log( MSE.Module.tabsPre().changeTab() ); // undefined
console.log( MSE.Module.tabsPre.changeTab() ); // Uncaught TypeError: MSE.Module.tabsPre.changeTab is not a function

I am trying to access changeTab() in the tabsPre object, but I don't seem to be able to. The last two console.log statements aren't giving me what I had hoped for. How can I do this?

Here's a JSFiddle: https://jsfiddle.net/xhb16qL6/

In the first console.log , I can see the function is there:

Any help or guidance on what I'm doing wrong would be great. I'm probably having a dumb day and can't see it.

Thanks, Mikey

Answer 1

ChangeTab返回传递给它的args，并且由于未传递任何参数而未定义打印，请尝试：

console.log(MSE.Module.tabsPre().changeTab("args"))  //"args"

Answer 2

console.log( MSE.Module.tabsPre() );

this logs the tabsPre object you are returning

console.log( MSE.Module.tabsPre().changeTab() );

this logs the result of MSE.Module.tabsPre().changeTab() which is undefined as you didn't pass an argument

console.log( MSE.Module.tabsPre.changeTab() );

this causes an error as MSE.Module.tabsPre is a function, and therefore you cannot access properties of it, as they don't exist

Answer 3

JavaScript dosen't check for the number of arguments.

Suppose I have a method, say add and defined like this:

function add(arg1, arg2){

}

I call this method like in 3 different ways, lets assume:

1) add(1,2)   // Works just fine

2) add(1,2,3) // My third argument is ignored.

3) add(1)    // 2nd expected parameter is taken to be `undefined`

Your problem is scenario number 3.

Your changeTab method expects ONE argument. If you dont pass any in your call to changeTab , you see undefined .

Answer 4

The changeTab () function expects an argument. The code then returns the argument passed once it's called. Since you didn't pass an argument, then undefined is returned.

Try passing an argument :)