Why wont my closure function work?

Question

The function below adds arguments within 1 paranthesis fine. For example, it computes addTogether(2,3) = 5 and addTogether(2,"3") = undefined.

However, it fails to compute addTogether(2)(3) = 5, instead giving me the error that "addTogether(...) is not a function". The closure function (return function(x)) is supposed to take into the second argument in addTogether(2)(3), and I'm lost on why it dos not work.

function addTogether() {
 if (typeof arguments[0] !== "number" || typeof arguments[1] !== "number") {
 return undefined;
 } //not harmful but not necessary
 var sum = 0;
 var num = arguments[0];
 if (arguments.length === 1) {
 //if only 1 argument in the original function...
 if (typeof arguments[0] !== "number") {
 return undefined;
 //returns undefined if arguments[0] isnt a number
 }
 return function(x) {
 if (typeof arguments[0] !== "number") {
 return undefined;
 //in the closure/2nd function, if first argument isnt a number then no sum will be provided
 } else {
 sum = num + x; //x = second given argument
 return sum;
 }
 };
 }
 if (typeof arguments[0] === "number" && typeof arguments[1] === "number") {
 for (var x = 0; x < arguments.length; x++) {
 if (typeof arguments[x] === "number") {
 sum += arguments[x];
 //add the argument[0] and [1] if both are number types, not string or array types or any other types
 } else {
 sum = undefined;
 }
 }
 return sum;
 }
 // the above "if" statement is rsponsible for achieving addTogether(2,3) = 5;
}
console.log(addTogether(2)(3));

• Sebastian Simon: if (typeof arguments[0] !== "number" || typeof arguments[1] !== "number") already causes addTogether(2) to return undefined.
• Felix Kling: ^ If you only pass a single argument, typeof arguments[1] !== "number" is true (because typeof arguments[1] will return "undefined").
• Barmar: The comment "Not harmful but not necessary" is wrong. It's not necessary, and it's harmful.
• Barmar: @Xufox Turn that into an answer.
• Felix Kling: FWIW, you could have easily find out why the function doesn't return a function if you used your browser's developer tools and set a breakpoint at the beginning of function. Take the time to learn how to debug JavaScript. It's invaluable.

Answer 1

If you want your function to work like addTogether(2)(3), this means that your addTogether must take an parameter and return a function. addTogether(2) this call will return a new function, and then call the returned function with the second parameter.

In your case when you compare

if (typeof arguments[0] !== "number" || typeof arguments[1] !== "number") 

and call the function with one argument, the second typeof arguments[1] !== "number" returns you true, because the second parameter is undefined, so it is not a number and your function returns undefined.

And in your code you can remove some conditions also. Because the above condition will already check them.

function addTogether() {
 if (typeof arguments[0] !== "number") { 
 return undefined;
 } 
 var sum = 0;
 var num = arguments[0];
 if (arguments.length === 1) { 
 return function(x) {
 if (typeof arguments[0] !== "number") {
 return undefined;
 
 } else {
 sum = num + x; 
 return sum;
 }
 };
 }
 if (typeof arguments[0] === "number" && typeof arguments[1] === "number") {
 for (var x = 0; x < arguments.length; x++) {
 if (typeof arguments[x] === "number") {
 sum += arguments[x];
 
 } else {
 sum = undefined;
 }
 }
 return sum;
 }
 
}
console.log(addTogether(2)(3));

• Python noob: Thank you for the help, but doesnt the (return function(x)) return that new function? Im confused how to fix this
• Suren Srapyan: @Pythonnoob Edited

Answer 2

if (typeof arguments[0] !== "number" || typeof arguments[1] !== "number") already causes addTogether(2) to return undefined.

Placing that if statement at the end of the function or turning that || into an && fixes it.