Question

I have an array of arbitrary values. I Wrote a function that transforms the array to an array of functions that return the original values, so instead of calling a[3], I will call a3.

Here is my code which does not work? code. It gives this error Cannot call method '1' of undefined.

var numToFun = [1, 2, { foo: "bar" }];
var numToFunLength = numToFun.length;

function transform(numTo) {
    for (var i = 0; i < numToFunLength; i++) {
        (function(num){
            numTo.unshift(function() {
                return num;
            });
        }(numTo.pop())) 
    }
}

var b = transform(numToFun);

console.log(numToFun);
console.log(b[1]());​

Answer 1

Others have already answered your question while I was writing mine but I will post it anyway – this may be somewhat easier to follow without all of those popping and unshifting:

function transform(numTo) {
    var r = [];
    for (var i = 0; i < numTo.length; i++) {
        r[i] = (function (v) {
            return function() {
                return v;
            }
        }(numTo[i]));
    }
    return r;
}

(I have also changed the hard-coded length from numToFunLength to numTo.length so the transform() function would work for other inputs than only the global numToFun variable.)

See DEMO.

UPDATE: even more elegant way to do it using the Sugar library:

function transform(array) {
    return array.map(function (v) {
        return function() {
            return v;
        }
    });
}

I like this syntax because it makes it more explicit that you want to map an array of values to an array of functions that return those values.

See DEMO.