I'm a TA and a student came in asking why the following code didn't swap the first 2 elements in an array and instead resulted as undefined. Here's the code the student showed me:
var swapFirstTwoElementsOf = function (a) {
a = [a[1],a[0]].concat(a.slice(2, a.length));
}
Why does this return undefined?
You need to return the variable. The local reference is reassigned, but the original variable a is not. You need to do something like
var swapFirstTwoElementsOf = function (a) {
return [a[1],a[0]].concat(a.slice(2, a.length));
}
var myArray = [0, 1, 2, 3];
myArray = swapFirstTwoELementsOf(myArray);
Previously, the function was evaluating to undefined because it wasn't returning anything.
var tmp = a[0]; a[0] = a[1]; a[1] = tmp; would swap the first two elements of the original array without the function needing to return anything - you can't reassign the original variable from inside the function, but you can modify the object it already references. (Though I realise the code was a question from a student so suggesting alternative methods may or may not be appropriate.)