Add 2 numbers

Let's start by creating a function called add that can accept 2 arguments and that returns their sum.

examples/js/add_2_numbers.js

function add(x, y) {
    return x+y;
}
console.log(add(2, 3));      // 5
console.log(add(-1, 1));     // 0

console.log(add(1, 1, 1));   // 2

We can use Node.js to run the code node add_2_numbers.js and the result will the the numbers seen next to each row.

As you can observe, the first two calls gave the correct result. The third one, where we passed 3 numbers return 2, which might be surprising to someone who does not know the implementation of the add function, but which is correct. The bigger issue is that it did not issue any warning or error complaining about the unused third parameter.

Add 3 numbers

We can try to create another add function that will accept 3 parameters and will return the sum of all 3. The results are next to the lines:

examples/js/add_3_numbers.js

function add(x, y) {
    return x+y;
}
function add(x, y, z) {
    return x+y+z;
}

console.log(add(2, 3));      // NaN
console.log(add(-1, 1));     // NaN

console.log(add(1, 1, 1));   // 3

JavaScript does not allow us to define the same function more than once with different signatures. It just silently overrides the previous function with the new function of the same name. So we would get the same results even if we remove the declaration with the 2 parameters.

What happens here is that in the first two cases, because we only passed 2 arguments, the third argument is undefined and when we add two numbers and undefined our result is NaN - not a number.

The third case, when we pass 3 arguments works as expected.

Clearly this is not the solution.

arguments, the object holding the passed arguments

The correct solution is to define a function with empty signature, as if it did not accept any parameters, and then to look at the arguments object that holds all the parameters.

We can iterate over the element using a for loop and add them all to an internal variable.

As you can see from the results this works even if the user did not pass any argument.

examples/js/sum_numbers.js

function sum() {
    var s = 0;
    for (var i=0; i < arguments.length; i++) {
        s += arguments[i];
    }
    return s;
}

console.log(sum(2, 3));         // 5
console.log(sum(-10, 1));       // -9
console.log(sum(1, 1, 1, 1));   // 4
console.log(sum());             // 0

For further details check out the documentation of arguments.