Question

var foo = function(){};
foo.prototype.value = 5;
foo.prototype.addValue = function(){ foo.value = 6; }
function bar(func)
{
func(); // I'm running the function!
}
bar(foo.addValue); // pass in the function 
alert(foo.value); // it's now 6!

Why is the no alert prompt when running this JavaScript code?

Answer 1

The correct code should be

var foo = function(){};
foo.prototype.value = 5;
foo.prototype.addValue = function(){ foo.value = 6; }
function bar(func)
{
    func(); // I'm running the function!
}
bar(foo.prototype.addValue); // pass in the function
alert(foo.value);

or

var foo = function(){};
foo.prototype.value = 5;
foo.prototype.addValue = function(){ foo.value = 6; }
function bar(func)
{
    func(); // I'm running the function!
}
bar(new foo().addValue); // pass in the function
alert(foo.value);

Since you are declaring a prototype called addValue, foo itself does not contain addValue. To refer to that function, you have to use foo.prototype.addValue. Note that in this case, foo contains both foo.prototype.value with a value of 5, and foo.value with a value of 6.

Alternatively, if the intention is to create a new object, then the object will inherit the addValue function, so calling new foo().addValue works.