从函数调用返回对象的函数会打印未定义

Question

var unique = function(){

var n=0;
return function(){

    return {

        inc : function(){
            n++;
            console.log(n);
        }

    };

 };
};

console.log(unique()().inc());

The code above prints 1 and then undefined what is the reason undefined gets printed ?

Answer 1

Because you asked to log the value returned by inc , and it doesn't return anything.

If you don't want to print anything,

console.log(unique()().inc());

should be

unique()().inc();

If you expect the new value of n to be printed,

inc : function(){ n++; console.log(n); }

should be

inc : function(){ n++; console.log(n); return n; }

Answer 2

you have 2 console.log, one inside inc (that prints 1), the other in the last line, (that prints undefined).

You can add return n; after the first console inside inc

    inc : function(){
        n++;
        console.log(n);
        return n;
    }