This week's book giveaway is in the OO, Patterns, UML and Refactoring forum. We're giving away four copies of Refactoring for Software Design Smells: Managing Technical Debt and have Girish Suryanarayana, Ganesh Samarthyam & Tushar Sharma on-line! See this thread for details.
Sri Its like this Look at the condition of the first for loop i<x.length x.length is the length of the x array which is clearly declared in the statement int  x=new int; so x.length is 3, ok, so the first for loop will run for 3 times for values of i=0, i=1, i=2 ok. Now the next for loop for(j=0;j<x[i].length;j++) here the condition is j shoulbe lessthan x[i].length,so when i is zero, the condition is j<x.length,ok Look at the line 5 from ur code it says x=new int ; so x.length is 4, so the condition is j<4; so the inner for loop will run for 4 times for j values of j=0, j=1, j=2 & j=3.now the value of i is zero.Now the sout will print 4 times ok. Now when i=1, the condition become j<x.length ie j<2 now the inner loop will run for 2 times for j values of j=0 and j=1.Now the sout will print 2 times. Now i=2, condition j<x.length ie j<5 so now the inner j loop will run for 5 times.now the sout will print 5 times. So totally when i=0; it prints 4 times when i=1; it prints 2 times when i=2; it prints 5 times So the answer is (5+2+4) 11 times. Hope this helps you.
Joined: Oct 26, 2004
Thanks Rangarajan that was a good explanation i got cleared now. Ofcourse i could find the clue earlier but this detailed flow cleared my doubts..thanks barry.