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JavaScript JavaScript Loops, Arrays and Objects Simplify Repetitive Tasks with Loops `do ... while` Loops

Mousa Zeidan
Mousa Zeidan
2,490 Points
Posted by Mousa Zeidan
Mousa Zeidan
2,490 Points

Why wouldn't while( guess !== randomNumber ) work in this case?

I tried this while condition and it just kept looping. From my understanding this means do (the code) only while your guess is not = to the random number generated. Logically this makes sense but it does not seem to work

am I missing something here :0?

Mousa Zeidan
Mousa Zeidan
2,490 Points
Mousa Zeidan
Mousa Zeidan
2,490 Points

Figured it out! I forgot to parseInt my guess!

while(parseInt(guess) !== randomNumber) works just fine. Is there any reason why we should stick to the way Dave taught it?

string
string
3,558 Points

If you use != it should work cos JS will just try and convert the string (guess) before comparing it to the number/int (randomNumber). But you used the strict inequality operator !==. It really is strict. It will try to compare their data type first even before looking at the values. And in this case, you're comparing a string and a number -- which evaluates to false. parseInt() makes them both actual numbers.

It's considered good practice to always use strict rules and syntax. Relying on JS automatic type conversion should be avoided as much as possible.

1 Answer

Dele Opeifa
Dele Opeifa
11,766 Points
Dele Opeifa
Dele Opeifa
11,766 Points

I did it this way and it worked perfectly.

do { guess = parseInt(prompt("I'm thinking of a number between 1 and 10. What is it?")); guessCount += 1; } while ( guess !== randomNumber )

I didn't have to add the extra variable that was used in the last step of the video. I think this is a good way to do it, but maybe I'm missing some potential bugs with this method?

The parseInt() is very important here, because prompt() will return a string, and your randomNumber variable is holding a number. Without parseInt, even if you technically type in the right number, it won't stop the loop, because it will be a string or characters and won't match the correct number.

Adrien Contee
Adrien Contee
4,875 Points
Adrien Contee
Adrien Contee
4,875 Points

I also did it this way without using the extra variable. Not sure why the extra false variable is even needed. But maybe later lessons will explain it better.