Manipulating Function

Question

Why is the output undefined even tho there is return statement.

function david(small_head) {
console.log(small_head);
}

function yusuff(brilliant, handsome) {
let grillo = brilliant + handsome;
return david(grillo);
}
console.log(yusuff(5,5));

Answer 1

Ibrahim Yusuf what you should do is: function david(xyz){ return xyz;
/*in this case, you are telling the function to return the result/value of xyz whenever the function is called*/
/* using console.log() instead of a return here would make the function undefined, cos it means no values is assigned to it*/};
function yusuf(b,h){
let g = b + h;
/* here, you are calling the function david inside of yusuf assigning the values/result from "g" to it as an argument, so a value would be returned*/
return david(g);
}
console.log(yusuf(5,5));
// i hope this helps.

Answer 2

Because the function david() didn't return any values. So that giveundefined.

Answer 3

Infinite, Vadivelan Do I need to return david() if it's already printed inside the function?

something like this:

function david(xyz) {
console.log(xyz);
}
david("Output");

even tho I didn't return david(), the output is still showing

Answer 4

Ibrahim Yusuf What do you want return from david()?

Answer 5

Vadivelan Just want to get the output of both david() and yusuff();

Answer 6

Infinite Thank you, That reall helps

Answer 7

There is no return in david().