js named function in parameter, can not access out of function

Question

Because the 5th edition of ECMAScript forbids use of arguments.callee() in strict mode.

So I decided do not use the callee, instead , I use a named function

The example in mdn

function factorial (n) {
    return !(n > 1) ? 1 : factorial(n - 1) * n;
}
[1,2,3,4,5].map(factorial);

become:

[1,2,3,4,5].map(function factorial(n) {
    return !(n > 1) ? 1 : /* what goes here? */ factorial(n - 1) * n;
});

This is a good idea, but I want to reuse the function factorial

DEMO

function d(x){return x;}

d(function a(){});
d(a);// this is not work, a is undefined(works in ie, failed in ff and chrome)

This is bother me, as I know, the scope in js is function-level, why the second a is undefined ?

jsfiddle DEMO

Can you be a little bit more clear? How are you not re-using the function factorial in the previous example? — Jonast92
– Jonast92, Commented Aug 22, 2013 at 16:19
The named function defined in parameter is not access-able out of the function.The previous example is show how I replace the callee by use the named function. — wener
– wener, Commented Aug 22, 2013 at 16:29

Ethan Brown · Accepted Answer · 2013-08-22 16:35:55Z

You're really asking two questions here. The first is easy:

[1,2,3,4,5].map(function factorial(n) {
    return !(n > 1) ? 1 : /* what goes here? */ factorial(n - 1) * n;
});

What you've got there works just fine. In other words, since you're naming your function (factorial), that identifier is available inside the function body, enabling recursion.

Your second question is a little trickier:

function d(x){return x;}

d(function a(){});
d(a);// this is not work, a is undefined(works in ie, failed in ff and chrome)

It is correct for a to be undefined. If you read the ECMAScript specification, section 13.2.1 clearly states that new execution context that's created includes the actual parameters (argument list). That is, a is defined within the execution context of d, not the caller. It looks like IE is expressing erroneous behavior.

Paul S. · Accepted Answer · 2013-08-22 16:18:00Z

1

The name of a function expression does not reference itself in the namespace it's defined in (IE bugs on this) but does reference itself within itself

[1].map(function foo() {return foo;});
// [function foo() {return foo;}]
foo;
// undefined

If you want a reference to the function so it can be re-used, create a reference to it, either go back to writing a function declaration (function as a statement), or set the function expression as a variable, and pass the variable as desired

var bar = function foo() {return foo;};
[1].map(bar);
// [function foo() {return foo;}]
foo;
// undefined
bar;
// function foo() {return foo;}

Note that inside the function named foo in this example, both foo and bar will point to the same function, unless bar is changed again later.

var bar = function foo() {return foo === bar;};
bar(); // true, bar is foo inside the function
// change value of bar, keep a reference to function foo
var temp = bar;
bar = 'something else';
temp(); // false, bar is no longer foo inside the function

edited Aug 22, 2013 at 16:18

answered Aug 22, 2013 at 16:10

Paul S.

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1 Comment

wener Over a year ago

This is the same as defined function direct,unless re-assigment the variable in run-time.This is no convenience as anonymous function.

Jonast92 · Accepted Answer · 2013-08-22 16:42:37Z

You say that your previous example doesn't work, but I have to disagree with you,

[1,2,3,4,5].map(function factorial(n) {
    return n == 0 ? 1 : factorial(n - 1) * n;
});

I tried this (yes I modified it a bit) using basic logging and it logged n-1 for every recursion indicating that the recursion works fine. I believe Paul explains just fine why.

I recommend, anyway, to change the way you're thinking here because imagine if you're finding factorial(100). You'd be doing

... * factorial(98 - 1) * 98 * 99 * 100

Imagine the overload. I recommend adding a optional parameter which is the sum of the currently evaluated factorials, your result is the same but the overload is way less.

[1,2,3,4,5].map(function factorial(n, sum) {
    var sum = sum || 1;
    return n == 0 ? 1 : factorial(n - 1, sum*n);
});

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