Cfrgtkky

public class Child{



    public static void main(String args){

        String x = new String("ABC");

        String y = x.toUpperCase();



        System.out.println(x == y);

    }

}

Output: true

So does toUpperCase() always create a new object?

toUpperCase() calls toUpperCase(Locale.getDefault()), which creates a new String object only if it has to. If the input String is already in upper case, it returns the input String.

This seems to be an implementation detail, though. I didn't find it mentioned in the Javadoc.

Here's an implementation:

public String toUpperCase(Locale locale) {

    if (locale == null) {

        throw new NullPointerException();

    }



    int firstLower;

    final int len = value.length;



    /* Now check if there are any characters that need to be changed. */

    scan: {

        for (firstLower = 0 ; firstLower < len; ) {

            int c = (int)value[firstLower];

            int srcCount;

            if ((c >= Character.MIN_HIGH_SURROGATE)

                    && (c <= Character.MAX_HIGH_SURROGATE)) {

                c = codePointAt(firstLower);

                srcCount = Character.charCount(c);

            } else {

                srcCount = 1;

            }

            int upperCaseChar = Character.toUpperCaseEx(c);

            if ((upperCaseChar == Character.ERROR)

                    || (c != upperCaseChar)) {

                break scan;

            }

            firstLower += srcCount;

        }

        return this; // <-- the original String is returned

    }

    ....

}