Java StringBuilder.subSequence() – Examples
In this tutorial, we will learn about the Java StringBuilder.subSequence() function, and learn how to use this function to find the subsequence of this StringBuilder sequence, with the help of examples.
subSequence(int start, int end)
StringBuilder.subSequence() returns a new character sequence that is a subsequence of this sequence whose start and end are specified as arguments.
Syntax
The syntax of subSequence() function is
subSequence(int start, int end)
where
| Parameter | Description |
|---|---|
| start | The index at which subsequence starts in StringBuilder. |
| end | The index at which subsequence ends in StringBuilder. |
Returns
The function returns CharSequence object.
Example 1 – subSequence(start, end)
In this example, we will take a StringBuilder initialized with a sequence "abcdefghijklm", and we shall find the subsequence that starts at index 3 and ends at index 7.
Java Program
public class Example {
public static void main(String[] args) {
StringBuilder stringBuilder = new StringBuilder("abcdefghijklm");
System.out.println("Original sequence : " + stringBuilder.toString());
int start = 3;
int end = 7;
CharSequence str = stringBuilder.subSequence(start, end);
System.out.println("Sub-sequence : " + str);
}
} Output
Original sequence : abcdefghijklm Sub-sequence : defg
Example 2 – subSequence(start, end) – end > length
In this example, we will take a StringBuilder initialized with a sequence "abcdefghijklm", and we shall find the subsequence that starts at index 3 and ends at index 15. Since, the end is greater than the length of this sequence, subSequence() throws java.lang.StringIndexOutOfBoundsException.
Java Program
public class Example {
public static void main(String[] args) {
StringBuilder stringBuilder = new StringBuilder("abcdefghijklm");
System.out.println("Original sequence : " + stringBuilder.toString());
int start = 3;
int end = 15;
CharSequence str = stringBuilder.subSequence(start, end);
System.out.println("Sub-sequence : " + str);
}
} Output
Original sequence : abcdefghijklm Exception in thread "main" java.lang.StringIndexOutOfBoundsException: start 3, end 15, length 13 at java.base/java.lang.AbstractStringBuilder.checkRangeSIOOBE(Unknown Source) at java.base/java.lang.AbstractStringBuilder.substring(Unknown Source) at java.base/java.lang.StringBuilder.substring(Unknown Source) at java.base/java.lang.AbstractStringBuilder.subSequence(Unknown Source) at java.base/java.lang.StringBuilder.subSequence(Unknown Source) at Example.main(Example.java:8)
Example 3 – subSequence(start, end) – start > end
In this example, we will take a StringBuilder initialized with a sequence "abcdefghijklm", and we shall find the subsequence that starts at index 7 and ends at index 3. Since, the end is greater than the length of this sequence, subSequence() throws java.lang.StringIndexOutOfBoundsException.
Java Program
public class Example {
public static void main(String[] args) {
StringBuilder stringBuilder = new StringBuilder("abcdefghijklm");
System.out.println("Original sequence : " + stringBuilder.toString());
int start = 7;
int end = 3;
CharSequence str = stringBuilder.subSequence(start, end);
System.out.println("Sub-sequence : " + str);
}
} Output
Original sequence : abcdefghijklm Exception in thread "main" java.lang.StringIndexOutOfBoundsException: start 7, end 3, length 13 at java.base/java.lang.AbstractStringBuilder.checkRangeSIOOBE(Unknown Source) at java.base/java.lang.AbstractStringBuilder.substring(Unknown Source) at java.base/java.lang.StringBuilder.substring(Unknown Source) at java.base/java.lang.AbstractStringBuilder.subSequence(Unknown Source) at java.base/java.lang.StringBuilder.subSequence(Unknown Source) at Example.main(Example.java:8)
Conclusion
In this Java Tutorial, we have learnt the syntax of Java StringBuilder.subSequence() function, and also learnt how to use this function with the help of examples.