Create a function that defines whether or not a word has even or odd number of characters?

Question

1.00/5 (2 votes)

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function helloYou(name){

}

/* Do not modify code below this line */
console.(helloYou('bob'), ←- Should return undefined')
console.log(helloYou('anna'), ←- Should return " Hello, Anna!"')

What I have tried:

I have no idea where to start. I am looking for someone that can explain functions briefly and how I can make a function read how whether not a word has an even amount of characters. If the word has an odd number of letters it needs to result in " undefined " If the word has an even number of letters it needs to result in the " Hello, Anna! " You can take a look at the code above to get a better idea of what it is I am asking.

Posted 20-May-20 9:10am

jd2426

Updated 20-May-20 10:19am

v2

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phil.o · Answer 1 · 2020-05-20T09:23:00

Solution 1

This allows you to get the length of a string:
String.length - JavaScript | MDN[^]

Then you can use this little bit trick to test whether a number is odd or even:

JavaScript

function isOdd(n)
{
   return n & 1 == 1;
}

The rest is just a matter of an if .. else block to implement the logic.

Posted 20-May-20 9:23am

phil.o

Comments

jd2426 20-May-20 15:32pm

Still semi clueless. I am new to coding. I am going to figure this out one way or another.

phil.o 20-May-20 16:00pm

In the helloYou function:
1. Get the length of the string parameter.
2. Test whether it is odd:
2a. If it is odd, return "undefined".
2b. Else, concatenate "Hello, ", the parameter, then "!", and return the result.

There is absolutely no difficulty nor trick in that. You just have to start writing and testing.

W∴ Balboos, *GHB* · Answer 2 · 2020-05-20T10:19:00

Solution 2

The only cure to clueless is getting a clue - and this is the place I used to learn javaScript.[^]

They make it easy - with examples that you can modify and see what happens.

Posted 20-May-20 10:19am

W∴ Balboos, GHB