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1. What will be the output of the program? String x = "xyz"; x.toUpperCase(); / Line 2 / String y = x.replace('Y', 'y'); y = y + "abc"; System.out.println(y);



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  • By: guest on 02 Jun 2017 01.26 am
    Line 2 creates a new String object with the value "XYZ", but this new object is immediately lost because there is no reference to it. Line 3 creates a new String object referenced by y. This new String object has the value "xyz" because there was no "Y" in the String object referred to by x. Line 4 creates a new String object, appends "abc" to the value "xyz", and refers y to the result.
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