Why does this code prints not only the argument passed to the function but also an undefined type.

Question

https://code.sololearn.com/W5PSTwvWlQd5/?ref=app

Answer 1

I'm sorry for misunderstanding but what I was saying is, when you use "console.log" it mean that you are printing something on console.

//Declaration of a function to print a variable name on console

function sayName (name) {
console.log(name)
}

The when you write

console.log(sayName('e'))

It means you are parsing a value "e" to a function so the statement starts
with a function call.

Then on console.log() there is no variable that's why it prints "undefined"

This means when you call

console.log(sayName ('e'))
console.log(sayName ('f'))

It will print

e
undefined
f
undefined

Answer 2

Yaroslav Vernigora thanks a lot for the explanation.

Answer 3

Ok. Thank you for being so nice. I'll refer to it on my web development journey.

Answer 4

Hi! because console.log after execution cannot return anything back.

function sayName(name){
console.log(name)
}

sayName('e')

Answer 5

The “undefined” value

The special value undefined also stands apart. It makes a type of its own, just like null.

The meaning of undefined is “value is not assigned”.

If a variable is declared, but not assigned, then its value is undefined:

let age;

alert(age);
// shows "undefined"

Technically, it is possible to explicitly assign undefined to a variable:

let age = 100;
// change the value to undefined

age = undefined;

alert(age); // "undefined"

…But we don’t recommend doing that. Normally, one uses null to assign an “empty” or “unknown” value to a variable, while undefined is reserved as a default initial value for unassigned things.

Answer 6

I'm advise you this site:

https://javascript.info/

Answer 7

you're welcome

Answer 8

Jessica Cristal Can you help me here please.

Answer 9

Yaroslav Vernigora thanks for checking out my code but i wanted to know why does my code prints not only 'e' but also 'undefined'

Answer 10

I wrote to you about it above. review again carefully my message.

Answer 11

Manoti (Challenge Accepted) first of all you are wrong for assuming what i wanted to do(because i was just playing around to learn) And you also didn't answer why my code( console.log(sayName('e')) prints undefined.

Answer 12

You declared a function with it's statement to print the name.
Also when calling the function you are also using a statement to print your function. You should just call the function

function sayName(name){
console.log(name)
}

sayName("e")

Answer 13

Mina Adel

Answer 14

Hello