Why does the console log undefined when running the third function?

Question

const addToTen = num => 10 + num;
const divideUs = (num1, num2) => num1/num2;
const printMyName = (test) => console.log(test);
console.log(addToTen(5));
console.log(divideUs(10,5));
console.log(printMyName('Chad'));
Console: 
treehouse:~/workspace$ node convert.js                                                                         
15                                                                                                             
2                                                                                                              
Chad                                                                                                           
undefined                                                                                                      
treehouse:~/workspace$

Aakash Srivastav · Answer

Hey friend  @Chadwick Savage (https://teamtreehouse.com/chadwicksavage)  , you have two mistakes here .   
First , you have already defined console.log() statement within your printMyName function , and you are using it outside too. 
Second , you have used = operator in between  function name and parenthesis , which should not be there
JavaScript
const printMyName(test) => console.log(test);

Just call the function with your desired name  like this printMyName('Chad')   
Hope it helps :)

Magnus Martin · Answer

Aakash is right there. I initially made the same mistake when copying my console.log statement. The printMyName function already logs to the console but returns nothing. So an extra undefined gets logged.

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Chadwick Savage

Chadwick Savage

Why does the console log undefined when running the third function?

2 Answers

Aakash Srivastav

Aakash Srivastav

Magnus Martin

Magnus Martin

Nelson J

Nelson J