So I'm trying to write this in a short way:
char letter;
while ( letter!='A' && letter!='B' && letter!= 'C... letter!= 'a'
&& letter !='b' && letter!=c)
Basically if the user does not input a letter between A and C, a while loop will run until A,B,C,a,b,c is inputted.
It should be in the form of
while(letter<'a' && letter > 'c')
but it didn't work apparently because if I inputted F, is it greater than 'C', but it is less than 'c', since char uses ACSII.
5 Answers
There are many ways to do this check.
char letter;
while (!(letter >= 'A' && letter <= 'C') && !(letter >= 'a' && letter <= 'c'))
Or you can use Character.toUpperCase or toLowerCase on the letter first, to remove half of the conditions.
Or, if the range of letters is small or non-contiguous, you could do:
char letter;
while ("ABCabc".indexOf(letter) == -1)
There are more ways of course.
Set letter to lower case and then check:
letter = Character.toLowerCase(letter);
while (letter < 'a' && letter > 'c') {
// ...
}
This way, even if the user enters an upper case letter, the check will work.
You need two conditions and a bit of de Morgan:
while(!((letter >= 'A' && letter <= 'C') || (letter >= 'a' && letter <= 'c')))
or good old regex
char input = 'B';
Pattern p = Pattern.compile("a|b|c", Pattern.CASE_INSENSITIVE);
Matcher m = p.matcher("" + input);
if (m.find())
{
System.out.println("found");
}
You can check with ASCII code of characters such as 65 for 'A' and 97 for 'a'. So you could do if (letter > 65) to check if letter greater than 'A'. Hope this helps. Updates : For checking if letter is only either of A,B,C,a,b,c, use this check : if (letter >= 65 && letter <= 67) || (letter >= 97 && letter <= 99) . Please mark as answer if this helps.
