Cannot figure the output

Question

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How does Duff’s device work?

int n=5;
int q=(n+3)/4;
switch(n%4)
{ 
  case 0:do{ n++;
  case 3:n++;
  case 2:n++;
  case 1:n++;}while(--q>0);
}
 cout<<n;

What will the value of n be? This is just the code snippet and the answer that is given is 10. Cannot see how?

make it compile and use a debugger to step through it and observe the variables changing — PlasmaHH, Nov 01 '12 at 12:57
Case labels are just labels, and `switch` is just a `goto`. Maybe that helps. — Kerrek SB, Nov 01 '12 at 12:58
Go step by step. How much are `n` and `q` before entering the switch? To what case the execution goes? What happens after that? — Dialecticus, Nov 01 '12 at 12:59
I don't think it will compile with a loop inside of a switch — kenny, Nov 01 '12 at 13:00
Well I have learned something here, Duff's Device. After all my c programming years I never managed to write anything as convoluted and unreadable as this. — Dave Richardson, Nov 01 '12 at 13:04

Dialecticus · Accepted Answer · 2012-11-01T14:30:52.133

1

Final value of n is 10. Before the switch n is 5, and q is 2. Switch goes to case 1. n is incremented 1 time in first iteration, and 4 more times in second. Finally n has the value 5+1+4 = 10.

edited Nov 01 '12 at 14:30

answered Nov 01 '12 at 13:07

Dialecticus

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Well, yes, now that the code is edited, `n` is indeed 10. – Dialecticus Nov 01 '12 at 14:30

Cannot figure the output

1 Answers1