Confusion areound returning function in case of one argument

JavaScript
#1 
function addTogether() {
let args = Array.from(arguments)
return args.some(n => typeof n !== "number") 
? undefined
: args.length > 1
? args.reduce((acc, n) => (acc += n), 0)
: n => (typeof n === "number" ? n + args[0] : undefined);
}

addTogether(2);

Would someone please help me understand how the above solution works to return a function if one argument is passed to addTogether function. thanks

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#2

This is definitely not the easiest code to read. The function is being returned here:

: n => (typeof n === "number" ? n + args[0] : undefined)

This is an anonymous arrow function that takes one argument (n).

The easiest way to read the return statement is to think of it as a bunch of if/else statements:

if (args.some(n => typeof n !== "number") === true) {
 return undefined;
}
else if (args.length > 1) {
  return args.reduce((acc, n) => (acc += n), 0);
}
else {
  return n => (typeof n === "number" ? n + args[0] : undefined);
}
1 Like
#3

Very well explained and I recalled the anonymous function now :slight_smile: thanks a lot