What will be the output of the program? class PassS { public static void main(String [] args) { PassS p = new PassS(); p.start(); } void start() { String s1 = "slip"; String s2 = fix(s1); System.out.println(s1 + " " + s2); } String fix(String s1) { s ?->(Show Answer!)

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1. What will be the output of the program? class PassS { public static void main(String [] args) { PassS p = new PassS(); p.start(); } void start() { String s1 = "slip"; String s2 = fix(s1); System.out.println(s1 + " " + s2); } String fix(String s1) { s1 = s1 + "stream"; System.out.print(s1 + " "); return "stream"; } }

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By: guest on 02 Jun 2017 01.26 am

When the fix() method is first entered, start()'s s1 and fix()'s s1 reference variables both refer to the same String object (with a value of "slip"). Fix()'s s1 is reassigned to a new object that is created when the concatenation occurs (this second String object has a value of "slipstream"). When the program returns to start(), another String object is created, referred to by s2 and with a value of "stream".

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