hi, there the output of following code is 10 and 40(i compiled and ran). i know why 10 got printed out, but i don't know why 40 got printed out(i thougt 20 would be the output). could some one explain to me?? public class Pass{ static int j=20; public static void main(String argv[]){ int i=10; Pass p = new Pass(); p.amethod(i); System.out.println(i); System.out.println(j); }
public void amethod(int x){ x=x*2; j=j*2; } }
Valentin Crettaz
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j is 20 2 * 20 is 40 Note that j has a class scope while the scope of i is within the method main ! HIH ------------------ Valentin Crettaz Sun Certified Programmer for Java 2 Platform
It prints out 40 because the j referenced in amethod() is the j declared as static int in the class, so whatever you do to it in amethod() will be visible in main(). i is not affected, of course, because x is local to amethod() and any changes to it are not visible in main().
------------------ Junilu Lacar Sun Certified Programmer for the Java� 2 Platform
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