Why Javascript is "typeerror"

Question

This is done to get the correct result, and the Chrome debugger says there were no exceptions:

var x = new Foo().bar().baz();

but it goes into space and never ends, and the Chrome debugger says that although bar () is executed correctly, it then throws a "typeerror" exception and never goes to the specified function when trying to call baz ():

var x = new Foo();
x = x.bar().baz();

It seems to me that they are functionally identical. Why do they behave differently?

Answer 1

My own question was answered - and boy, I feel stoopid! I'm sorry. A short answer to the question "Why do they behave differently?" - they are not .

, , , ... ! , "baz()", , ell ( "l" ) .

, . ( , - .)

, , . Chrome "typeerror", Firebug " ", .

: : -)

Answer 2

, Chrome:

function Foo(){}

Foo.prototype.bar = function() {
  return this;
}

Foo.prototype.baz = function() {
  return 'baz';
}

var x = new Foo().bar().baz();
console.log(x); // baz

(Firefox, IE, Chrome).

:

var x = new Foo();
x = x.bar().baz()
console.log(x); // baz

.

Answer 3

You need brackets:

var x = (new Foo()).bar().baz();

Otherwise, the statement newapplies to the return value Foo().bar().baz(), all of which will be executed before an attempt is made to create an instance.

Update . I was informed that this, as they say, is wrong. But I think the comments are interesting, so I am not deleting the answer.