javascript not nested function scope widening

Question

function foo(b)
{
    return cool
    (
        function(x)
        {
            if(x)
            {
                b(x);
            }
        }
    );
}

where cool is a function that takes in a function. This code works well enough. How can I make this work though?

function bar(x)
{
    if(x)
    {
        b(x);
    }
}
function foo(b)
{
    return cool(bar);
}

I want to do this because bar is a frequently used function from functions that are like foo. Is there any way to open up the scope further so that bar can see b from foo?

Unfortunately I don't have access to the cool function. Oh well it was worth asking. — No_name
– No_name, Commented Jan 14, 2015 at 2:58
Ah, I misread the code, please disregard that. I thought you were returning a function called cool. Take a look at Daniel's answer below. — Qantas 94 Heavy
– Qantas 94 Heavy, Commented Jan 14, 2015 at 2:59

Daniel · Accepted Answer · 2015-01-14 02:59:14Z

3

Wrap bar in a function that takes b as an argument: i.e.

function baz(b)
{
    return function(x){
        if(x)
        {
            b(x);
        }
    }
}

function foo(b)
{
    return cool(baz(b));
}

answered Jan 14, 2015 at 2:59

Daniel

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