JavaScript converting an array to array of functions

Question

Hello I'm working on a problem that requires me to change an set array of numbers into an array that returns the original numbers as a function. So we get a return of a2 instead of a[2].

I dont want the answer I just need a hint. I know i can loop through the array and use .pop() to get the last value of the array, but then I dont know how to convert it to a function from there. any hints?

• rlemon: if this is homework it should be tagged as such. it is ok if it is. just make sure you tag it
• mechanical_meat: and the reason for that is: when questions are tagged w/ "homework" tag, users of the site know that coding up a full solution is not as desirable as pointing you in the right direction
• Will Buck: Good point, I didn't really consider that. Thanks for the tip, hopefully my answer isn't going to far and is adding some knowledge! (let me know if it isn't)

Answer 1

var numToFun = [1, 2, 3];
var numToFunLength = numToFun.length;
for (var i = 0; i < numToFunLength; i++) {
 (function(num){
 numToFun.unshift(function() {
 return num;
 });
 }(numToFun.pop())) 
}

DEMO

basically it pops out a number from the last, builds a function with that number returned, and put back into the first of the array. after one full cycle, all of them are functions.

here's the catch: how this works, it's up to you to research

why the loop does not look like the straightforward pop-unshift:

for (var i = 0; i < numToFunLength; i++) {
 numToFun.unshift(function() { //put into first a function
 return numToFun.pop() //that returns a number
 });
}

and why i did this: (HINT: performance)

var numToFunLength = numToFun.length;

Answer 2

There's three important steps here:

• Extract the number value from the array. Within a loop with an iterator of i, it might look like this:

  var num = numArray[i];

• This is important, because i will not retain its value that it had when you created the new function - it'll end up with the last value it had, once the for loop is finished. The function itself might look like this:

  function() { return num; }

• There's no reference to i any more, which is important - to understand better, read about closures. The final step would be to add the new function to the array of functions that you want.

...and you're done!

• Stoive: I should add, that this homework task seems to deliberately target the common mistake regarding iterators and closures. Thumbs up to the teacher/lecturer/tutor/whatever!

Answer 3

EDIT: See other's answers for good explanations of how to do this right, I will fix mine also though

As others have pointed out, one of the tricky things in javascript that many struggle with (myself included, obviously) is that scoping variables in javascript is dissimilar to many other languages; scopes are almost purely defined by functions, not the {} blocks of, for example, a for loop, as java/C would be.

So, below you can see (and in other answers here) a scoping function can aid with such a problem.

var numArray = [12, 33, 55];
var funcArray = [];
var numArrLength = numArray.length; // Don't do this in for loop to avoid the check multiple times
for(var j=0; j < numArrLength; j++) {
 var scopeMe = function() {
 var numToReturn = numArray[j];
 console.log('now loading... ' + numToReturn);
 var newFunc = function() {
 return numToReturn;
 };
 return newFunc;
 }();
 funcArray.push(scopeMe);
};
console.log('now me');
console.log(funcArray);
console.log(funcArray[0]());
console.log(funcArray[1]());
console.log(funcArray[2]());
console.log(funcArray[1]()); // To ensure it's repeatable

EDIT my old bad answer below

What you'll want to do is something like

var funcArray = [];
for(...) {
 var newFunc = function() {
 return numArray.pop();
 }
 funcArray.push(newFunc); 
}

The key here is that functions in javascript can be named variables, and passed around as such :)

• Stoive: This probably won't work - i will be outside the newFunc closure, and return the last item every time.
• Will Buck: perhaps a more generic numArray.pop(); would be better? not sure if ordering is important to the problem or not :)
• Will Buck: I'm not really sure I deserved downvotes for that though? Can I get feedback for downvotes?
• Joseph: try testing in jsFiddle. you will see that the values aren't the same as you think they'd be.
• Stoive: The edited version still won't work - it'll return different values each time the function is called.