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Chadwick Savage
11,919 PointsWhy does the console log undefined when running the third function?
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$
2 Answers
Aakash Srivastav
Full Stack JavaScript Techdegree Student 11,638 PointsHey friend Chadwick Savage , 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
const printMyName(test) => console.log(test);
Just call the function with your desired name like this printMyName('Chad')
Hope it helps :)
Magnus Martin
44,123 PointsAakash 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.
Nelson J
7,411 PointsThanks I was having the same problem.