[FIXED] What are some ways to avoid String.substring from returning substring with invalid unicode character?

Issue

Recently, only I notice that, it is possible for substring to return string with invalid unicode character.

For instance

public class Main {

    public static void main(String[] args) {
        String text = "🥦_Salade verte";

        /* We should avoid using endIndex = 1, as it will cause an invalid character in the returned substring. */
        // 1 : ?
        System.out.println("1 : " + text.substring(0, 1));

        // 2 : 🥦
        System.out.println("2 : " + text.substring(0, 2));

        // 3 : 🥦_
        System.out.println("3 : " + text.substring(0, 3));

        // 4 : 🥦_S
        System.out.println("4 : " + text.substring(0, 4));
    }
}

I was wondering, when trimming a long string with String.substring, what are some good ways to avoid the returned substring from containing invalid unicode?

Solution

char obsolete

The char type has been legacy since Java 2, essentially broken. As a 16-bit value, char is physically incapable of representing most characters.

Your discovery suggests that the String#substring command is char based. Hence the problem shown in your code.

Code point

Instead, use code point integer numbers when working with individual characters.

int[] codePoints = "🥦_Salade".codePoints().toArray() ;

[129382, 95, 83, 97, 108, 97, 100, 101]

Extract the first character’s code point.

int codePoint = codePoints[ 0 ] ;

129382

Make a single-character String object for that code point.

String firstCharacter = Character.toString( codePoint ) ; 

🥦

You can grab a subset of that int array of code points.

int[] firstFewCodePoints = Arrays.copyOfRange( codePoints , 0 , 3 ) ;

And make a String object from those code points.

String s = 
    Arrays
        .stream( firstFewCodePoints ) 
        .collect( StringBuilder::new , StringBuilder::appendCodePoint , StringBuilder::append )
        .toString();

🥦_S

Or we can use a constructor of String to take a subset of the array.

String result = new String( codePoints , 0 , 3 ) ;

🥦_S

See this code run live at IdeOne.com.

Answered By - Basil Bourque