CODEWITHSUNDEEP
javascript
user1386284
user1386284

Reputation: 95

Javascript calling function expression

It makes sense by calling the function this way:

print(square(5));  
function square(n){return n*n}

But why the following calling doesn't work?

print(square(5));  
square = function (n) {  
  return n * n;  
}

What's the solution if we insist to use the format of "square = function (n)"?

Upvotes: 8

Views: 7134

Answers (6)

Alnitak
Alnitak

Reputation: 339917

"normal" function declarations are hoisted to the top of the scope, so they're always available.

Variable declarations are also hoisted, but the assignment doesn't happen until that particular line of code is executed.

So if you do var foo = function() { ... } you're creating the variable foo in the scope and it's initially undefined, and only later does that variable get assigned the anonymous function reference.

If "later" is after you tried to use it, the interpreter won't complain about an unknown variable (it does already exist, after all), but it will complain about you trying to call an undefined function reference.

Upvotes: 12

shashwat gupta
shashwat gupta

Reputation: 9

var s=function ()
	{

		console.log("s");
		alert("function expression with anomious function");

		
	}

s();

var otherMethod=function ()
	{

		console.log("s");
		alert("function expression with function name");

		
	}

otherMethod();

Upvotes: -1

shashwat gupta
shashwat gupta

Reputation: 9

    var s=function ()
	{
		console.log("hi there");
        document.write("function express called");
		alert("function express called");

		
	}

	s();

Upvotes: 1

Greg Hewgill
Greg Hewgill

Reputation: 993901

In the second case, square is a regular variable subject to (re)assignment. Consider:

square = function (n) {  
  return "sponge";  
}  
print(square(5));  
square = function (n) {  
  return n * n;  
}

What would you expect the output to be here?

Upvotes: 0

Esailija
Esailija

Reputation: 140228

In function expression you are using the function like any other value, would you expect:

print(a);
var a = 5

to work? (I'm not really asking)

Upvotes: 0

gdoron
gdoron

Reputation: 150273

You need to change the order, you use the variable before it was declared and assigned:

square = function (n) {//Better use "var" here to avoid polluting the outer scope
  return n * n;  
}  
print(square(5));

Correct way with var :

var square = function (n) { // The variable is now internal to the function scope
  return n * n;  
}  
print(square(5));

Upvotes: 0

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