Question

Can you explain me why does the second call of fn gives an error? The code is below.

function Test(n) {
  this.test = n;

  var bob = function (n) {
      this.test = n;
  };

  this.fn = function (n) {
    bob(n);
    console.log(this.test);
  };
}

var test = new Test(5);

test.fn(1); // returns 5
test.fn(2); // returns TypeError: 'undefined' is not a function

Here’s a JSfiddle that reproduces the error http://jsfiddle.net/KjkQ2/

Answer 1

Your bob function is called from the global scope. Thefore, this.test is pointing at a global variable named test which is overwriting the variable you created. If you run console.log(window.test), you’ll what’s happening.

For your code to behave as intended, you would need one of the following

function Test(n) {
  this.test = n;

  // If a function needs 'this' it should be attached to 'this'       
  this.bob = function (n) {
      this.test = n;
  };

  this.fn = function (n) {
    // and called with this.functionName
    this.bob(n);
    console.log(this.test);
  };
}

OR

function Test(n) {
  this.test = n;

  var bob = function (n) {
      this.test = n;
  };

  this.fn = function (n) {
    // Make sure you call bob with the right 'this'
    bob.call(this, n);
    console.log(this.test);
  };
}

OR closure based objects

// Just use closures instead of relying on this
function Test(n) {
  var test = n;

  var bob = function (n) {
      test = n;
  };

  this.fn = function (n) {
    bob(n);
    console.log(test);
  };
}