Discussion :: Java.lang Class
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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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Answer : Option C
Explanation :
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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