<script> !function add_them(a,b) { return a+b;} (9,4) console.log(add_them()); </script> Question:
it shows: ReferenceError: add_them is not defined, what is the problem? and how to make it show the right result: 13?
- @BenjaminGruenbaum Thanks for the response.. I found that out, so deleted the comment..karthikr– karthikr2013-06-21 02:03:57 +00:00Commented Jun 21, 2013 at 2:03
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3 Answers
You seem to want a simple function declaration:
function add_them(a,b) { return a+b;} console.log(add_them(9,4)); If you wanted to use an immediately invoked function expression, it might look like this:
console.log(function add_them(a,b) { return a+b; } (9,4)); // \ expression / ^^^^^ // invocation however the identifier (add_them) is not reusable outside (after) of the function expression.
Also notice that when your IEFE is supposed to yield some result, the (little) advantage of !function(){}() over (function () {})()? is neglected since the result gets negated.
2 Comments
user2294256
thanks, why i can not do this way: console.log(!function add_them(a,b) { return a+b; } (9,4)); !function () {}(); will also invoke the function, right?
Benjamin Gruenbaum
@user2294256 The ! operator actually changes the value where the ( just forces it into an expression. You're performing negation on 13 which returns 'false'.
You're confusing the browser (well, it's technically correct according to the spec).
Try another trick to make it treat the function as an expression:
(function add_them(a,b) { return a+b;})(9,4) // 13 Or simply:
console.log((function add_them(a,b) { return a+b;})(9,4)) // logs 13 If you're just trying to fixate the parameters you can .bind the function to values (which is what it seems to me you're trying to do):
var add_them = (function(a,b) { return a+b;}).bind(null, 9,4) console.log(add_them());// logs 13 answered Jun 21, 2013 at 2:02
Benjamin Gruenbaum
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The problem is that immediately invoked function expressions are function expressions.
In JavaScript functions may be either declarations or expressions. For instance, consider:
(function (a, b) { return a + b; }) // This is an expression function (a, b) { return a + b; } // This is a declaration How to distinguish a function expression from a function declaration? If the function is used in place where an expression is expected it automatically becomes an expression. Function expressions can be invoked immediately.
On the other hand a function declaration is also easy to spot because declarations in JavaScript are hoisted. Hence the function can be invoked before it appears in the program. For example, consider:
console.log(add(2, 3)); function add(a, b) { return a + b; } Now the problem with your code is that you're immediately invoking a function expression. However because it's an expression and not a declaration you can't use it like a normal function. You need to declare a function before you can use it.
Do this instead:
add_them(9, 4); console.log(add_them()); function add_them(a, b) { return a + b; } If you want to fix the parameters of the function then you may use bind as follows:
var add_9_4 = add_them.bind(null, 9, 4); console.log(add_9_4()); function add_them(a, b) { return a + b; } Hope that helps.
