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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T18:47:35+00:00 2026-05-15T18:47:35+00:00

In Java I am using the substring() method and I’m not sure why it

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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
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  1. Editorial Team
    Editorial Team
    2026-05-15T18:47:36+00:00Added an answer on May 15, 2026 at 6:47 pm

    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.

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