How to Print the output of a function using console.log (Javascript)

Question

The following code is meant print the namestring with a name. however, it's not correct.

var nameString = function (name) {
    return "Hi, I am" + " " + name.

}
nameString("Amir")
console.log(nameString)

What am I not implementing/doing wrong that stops it from displaying the string as well as a name? thanks.

You have to actually return the result of the function somewhere, just calling it isn't enough! — adeneo, Jul 17 '14 at 08:39
`nameString` is function definition, this will print function. `nameString()` will execute the function, this will print the output — Saahithyan Vigneswaran, Jul 21 '18 at 10:22

score 7 · Accepted Answer · answered Jul 17 '14 at 08:44

7

First mistake in your code is in the line

 return "Hi, I am" + " " + name.

remove fullstop or just concatenate it as below

 return "Hi, I am" + " " + name+"."

and then write

console.log(nameString("Amir"));

check it here fiddle

answered Jul 17 '14 at 08:44

Roshan

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the fullstop was throwing the entire code off, thanks for pointing it out, and now it's fixed. Beginners mistake that i will remember for next time, thanks again! – user3848245 Jul 17 '14 at 08:46
I still have to wait a few minutes, before I can accept an answer. I'll do it as soon as I can. – user3848245 Jul 17 '14 at 08:50

Mritunjay · Answer 2 · 2014-07-17T08:43:27.257

2

you will have to call that function.

console.log(nameString("Amir"));

If you will say console.log(nameString) it will just print the value of varibale nameString which is a function.

There is a . in your function in last of return statement, remove that.

edited Jul 17 '14 at 08:43

answered Jul 17 '14 at 08:38

Mritunjay

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7
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thank you as well for pointing out the full stop in the return statement, beginners mistake and it won't happen again. thanks! – user3848245 Jul 17 '14 at 08:47

score 2 · Answer 3 · answered Jul 17 '14 at 08:38

2

console.log(nameString())

just forgot to CALL the function. and pass params

console.log(nameString('anyString'))

answered Jul 17 '14 at 08:38

Gena Moroz

929
5
9

Govind Singh · Answer 4 · 2014-07-18T05:43:43.333

1

nameString is a method and you are not passing argument to it

console.log(nameString("Amir"));

or

var str=nameString("Amir");
console.log(str);

and also remove .

edited Jul 18 '14 at 05:43

answered Jul 17 '14 at 08:38

Govind Singh

15,282
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106

score 0 · Answer 5 · answered Oct 02 '15 at 14:34

0

var nameString = function(name) {
  return "Hi, I am" + " " + name
  nameString("Sophie");
}
console.log(nameString("Sophie"));

answered Oct 02 '15 at 14:34

SophieA.

1

score 0 · Answer 6 · answered Oct 09 '21 at 14:56

You can use template literal, which makes the code more readable and makes your job easier. You don't have to think about spaces and there are fewer inverted commas.

var nameString = function (name) {
  return `Hi, I'm ${name}.`;
};
nameString("Amir");
console.log(nameString("Amir"));

How to Print the output of a function using console.log (Javascript)

6 Answers6

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