Why is "out of range" not thrown for 'substring(startIndex, endIndex)'

Question

In Java I am using the substring() method and I'm not sure why it is not throwing an "out of index" error.

The string abcde has index start from 0 to 4, but the substring() method takes startIndex and endIndex as arguments based on the fact that I can call foo.substring(0) and get "abcde".

Then why does substring(5) work? That index should be out of range. What is the explanation?

/*
1234
abcde
*/
String foo = "abcde";
System.out.println(foo.substring(0));
System.out.println(foo.substring(1));
System.out.println(foo.substring(2));
System.out.println(foo.substring(3));
System.out.println(foo.substring(4));
System.out.println(foo.substring(5));

This code outputs:

abcde
bcde
cde
de
e
     //foo.substring(5) output nothing here, isn't this out of range?

When I replace 5 with 6:

foo.substring(6)

Then I get error:

Exception in thread "main" java.lang.StringIndexOutOfBoundsException:
    String index out of range: -1

Answer 1

According to the Java API doc, substring throws an error when the start index is greater than the Length of the String.

IndexOutOfBoundsException - if beginIndex is negative or larger than the length of this String object.

In fact, they give an example much like yours:

"emptiness".substring(9) returns "" (an empty string)

I guess this means it is best to think of a Java String as the following, where an index is wrapped in |:

|0| A |1| B |2| C |3| D |4| E |5|

Which is to say a string has both a start and end index.

Answer 2

When you do foo.substring(5), it gets the substring starting at the position right after the "e" and ending at the end of the string. Incidentally, the start and end position happen to be the same. Thus, empty string. You can think of the index as being not an actual character in the string, but a position in between characters.

        ---------------------
String: | a | b | c | d | e |
        ---------------------
Index:  0   1   2   3   4   5