Call a function in a function

Question

There is such a structure (which it would be desirable not to change dramatically):

var wrapper1 = function () { var wrapper2 = function () { return Math.PI; }; };

I tried using var caller = new wrapper1; call caller.wrapper2() , but returns undefined.
Tell me how to be, please. There is no way to understand the op in js.

Accepted Answer · 2017-02-01T19:40:57

Replace var wrapper2 = ... with this.wrapper2 = ... When you write var inside a function, this variable (which refers to the function) is visible only inside the function. If you access this (in the case of javascript when creating a new instance of an object through new somefunction() this inside this constructor function itself points to the created object. Then through this.wrapper2 = ... you create a new field in the object and put in its own function.

Not so, then if there is such a structure, then everything will collapse.
@VostokSisters you have a variable wrapper3 that gets the value returned by the wrapper2 function, not this.wrapper2.
replace wrapper2 () with this.wrapper2 () in the body of the last anonymous function
On the caller object or on the wrapper1 function wrapper1 ?

user207618 · Answer 2 · 2017-02-01T19:42:09

Of course, because wrapper2 is a private variable; it exists only during the execution of the function (or in the closure, which is not there).
Use this :

 var wrapper1 = function () { this.wrapper2 = function () { return Math.PI; }; } var caller = new wrapper1; console.info(caller.wrapper2());

Not so, then if there is such a structure, then everything will collapse.
In other words, wrapper2() should be visible in the wrapper1 osprey.
@VostokSisters, and who's stopping you from calling this: return this.wrapper2();
Yes, no one, but then it does not work with "use strict" ...

Call a function in a function

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