javascript未嵌套功能范围扩大

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 ?

Answer 1

Wrap bar in a function that takes b as an argument: ie

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

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