Java - How does the String constructor convert the char array to String when there are numbers as arguments?

Anshul V. Kumar 0
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I am confused as to how the Java String class default constructor converts the char array into String.

char c[]= {'H', 'E', 'L', 'L', 'O'};
String s= new String (c, 0,3);
System.out.println(s) ;

Gives output as - HEL

In this case, the last character is taken as endIndex - 1.

char c[]= {'H', 'E', 'L', 'L', 'O'};
String s= new String (c, 1,3);
System.out.println(s) ;

Gives output as - ELL

But, in this case, the last character is taken as endIndex only.

What's happening here?

default constructor string java arrays
1 Answers

The third parameter is count not endIndex, it specifies how many characters are copied from the input. From the docs:

public String​(char[] value, int offset, int count)

Allocates a new String that contains characters from a subarray of the character array argument. The offset argument is the index of the first character of the subarray and the count argument specifies the length of the subarray. The contents of the subarray are copied; subsequent modification of the character array does not affect the newly created string. [emphasis mine]

knittl0
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