Issue
Is there a nicer way of converting a number to its alphabetic equivalent than this?
private String getCharForNumber(int i) {
char[] alphabet = "ABCDEFGHIJKLMNOPQRSTUVWXYZ".toCharArray();
if (i > 25) {
return null;
}
return Character.toString(alphabet[i]);
}
Maybe something than can deal with numbers greater than 26 more elegantly too?
Solution
Just make use of the ASCII representation.
private String getCharForNumber(int i) {
return i > 0 && i < 27 ? String.valueOf((char)(i + 64)) : null;
}
Note: This assumes that i
is between 1
and 26
inclusive.
You'll have to change the condition to i > -1 && i < 26
and the increment to 65
if you want i
to be zero-based.
Here is the full ASCII table, in case you need to refer to:
Edit:
As some folks suggested here, it's much more readable to directly use the character 'A'
instead of its ASCII code.
private String getCharForNumber(int i) {
return i > 0 && i < 27 ? String.valueOf((char)(i + 'A' - 1)) : null;
}
Answered By - adarshr
Answer Checked By - Mildred Charles (JavaFixing Admin)