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JavaScript

Can't figure this one out

I am trying to make a program that takes user input of a name and turns their input into regular names, hyphenated names, nicknames, prefixes, and suffixes. All of it has gone well so far, except for nicknames. I tell the user to put their nicknames in quotes (single or double) and I will be able to print what they typed in the quotes. Example:

Input: Dwayne 'The Rock' Johnson Output: Regular Names: Dwayne, Johnson; Nicknames: 'The Rock'

I have made a loop to iterate through all the characters in the input and try and find nicknames, here it is:

 // get nicknames

    for (let a = 0; a < char.length; a ++) {

      if (char[a] == "'" || char[a] == '"') {

        joinarr.push(char[a]);

        for (let b = a + 1; true; b ++) {

          if (char[b] == char[a]) {

            joinarr.push(char[b]);
            break;
          }
          joinarr.push(char[b]);
        }
      }
    }

Notice that the loop with 'a' as a counter loops through all the characters. In the loop, it tests If char[a] is a single or double quote. If it is, it adds that quote to the joinarr array. It then enters a loop with a true condition that uses 'b' as a counter, with a value of a + 1. It then tests if char[b] is equal to char[a]. If it is, that would mean that char[b] is the end quote for that nickname, so it adds the end quote to the array and breaks the loop, otherwise it just adds char[b] to the array. My problem is that when I run this, my program crashes, telling me that the true loop never stops which is strange because I always type the corresponding end quote. Been at this a couple days, anyone any ideas?

1 Answer

Steven Parker
Steven Parker
220,425 Points

The first time the "a" loop encounters a quote, the "b" loop will find a matching one. But the next time the "a" loop encounters a quote it will be the ending one and there will be no match.

You probably want to advance index "a" before you break out of the "b" loop ("a = b + 1"). Then the outer scan will continue where the inner one left off.

But there's also no reason to allow the "b" loop to be unlimited and crash on a malformed input. A sensible limit would be to stop if the end of the string is reached ("b < char.length").

You are right, it probably would be more sensible to make a limit rather than have it crash. Also thank you for the fix. I spent so much time focusing on b's index that I didn't think about a being stuck on the first quote. Thanks Steven