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Options : 1)The x value is 20 2)The x value is 30 3)The x value is 400 4)The x value is 600 Ans : 1)The x value is 20 I feel that the ans is 3)The x value is 400, Because in the method, y is 20 and 20*20 will give 400 Please explain me..
I know the X value only changed in the Method,But don't how to change the X value out the method. Who can explain it ,thanks!
help you means help me
Joined: Jan 08, 2002
Originally posted by Arsho, Ayan: Hi Sonir : The main method just cannot access instance variable without creating an instance.The only 'x' variable accessible is local variable. so it prints 20. Hope that helps!! -A
I think this sentence "MyTest ta = new MyTest();" creating an instance, doesn't it?
Hi Sonir, There are three variables named 'x' in the example and all three have a different scope. int x = 30; is associated with each instance but is out of scope as main() is a static method. Line int x = y*y; in method 'Method()' defines another variable 'x' whose scope is limited to to braces of method 'Method()'. When the code tries to print value of 'x' the only x it can see (or is accessible) is the one initialised as int x = 20 and hence the answer. Rgds., Bhushan