Java substring endindex-1

Question

I was using substring in a project, earlier I was using endindex position which was not creating the expected result, later I came to know that java substring does endindex-1. I found it not useful. So why Java is doing endindex-1 instead of plain endindex ?

My code is as follows.

String str = "0123456789";
String parameter1 = str.substring(0,4);
String parameter2 = str.substring(5,8);`

Solution 1

It has a number of advantages, not least that s.substring(a,b) has length b-a.

Perhaps even more useful is that

s.substring(0,n)

comes out the same as

s.substring(0,k) + s.substring(k,n)

Solution 2

The javadoc explains why, by having it this way

endIndex-beginIndex = length of the substring

The parameters are therefore defined as

beginIndex - the beginning index, inclusive. endIndex - the ending index, exclusive.

A use case would be when you do something like

int index = 3;
int length = 2;
str.substring(index,index+length); 

Its just easier to programatically do things when uses the endIndex in the exclusive way

Solution 3

Java always uses the begin parameter as inclusive and the end parameter as exclusive to determine a range.

[0:3[ <=> 0, 1, 2
[4:6[ <=> 4, 5

you will find this behaviour several time within the API of java: List, String, ...

it's defined years ago within the java spec. and imho this is awesome and very useful. So with this definition you can make this.

String x = "hello world";
String y = x.substring(0, x.length() - 6); // 6 = number of chars for "world" + one space.
StringBuilder builder = new StringBuilder(y);
builder.append(x.subSequence(5, x.length())); // 5 is index of the space cause java index always starts with 0.
System.out.println(builder.toString());

Otherwise you always have to calculate to get the real last char of a String.