codePointCount() in java problem

Question

https://code.sololearn.com/cw3uFPz49zmH/?ref=app
According to ex.
Why is the final(end) index 5 or does it not give an error in 5? While the string is of 5 characters.
0=H 1=e 2=l 3=l 4=o 5=? In copePointCount() method
While the 5th index returns an error upon printing .In charAt() method

Answer 1

The syntax of codePointCount() is
codePointCount(int beginIndex, int endIndex)
Here if endIndex will be greater than size of String then it will give arrayIndexOfBoundException. But in your case endIndex is 5 which is equal to length of String that's why not giving exception. When you change to 6 then it gives exception.
To know more check this :-
https://www.w3resource.com/java-tutorial/string/string_codepointcount.php

Answer 2

ABHISHEK PATEL Because charAt depends on index which is start from 0. So in your case the last index is 4 but you are accessing 5th index.
The index number starts from 0 and goes to n-1, where n is length of the string. It returns StringIndexOutOfBoundsException if given index number is greater than or equal to this string length or a negative number.
Check this link.
https://www.javatpoint.com/java-string-charat

Answer 3

Because the index AFTER the last Character throw an IndexOutOfBoundsException at the charAt method. Every method have his own Parameter and edges. Mean boundary conditions. for more detail refer the Documentation.

Answer 4

But why charAt() method giving error. if length 5 according to you