Could anyone please explain why the output of this code is 4 when x = 0 ( i thought if x is 0, which applies the the case 0: and y suppose to be 3 and i'm wrong in this case.
#include <stdio.h> int main() { int x = 0; int y; switch (x) { case 0: y = 3; case 1: y = 4; break; default: y = 5; break; } printf("%d", y); getchar(); return 0; } After switch(x) operation, execution jumps to the corresponding case N statement first. Then it executes code after that statement until a) it find the end of the switch(x) boundary defined by brackets ({}) or b) it finds a break statement inside the switch, which ends that switch.
In your case it went through y=3 down to case 1, down to y=4 and then ended at the break.
You commented out your break; statement in your first case statement!
So just write this:
(Otherwise when x is 0 or 1 y gets assign to 3 and then 4)
case 0: y = 3; break; A switch statement goes to the next break; if there is missing one for a case statement! An if you forgot all of them :D then it stops at then end of the switch statement and all lines get's executed!
Because the break is commented out:
switch (x) { case 0: y = 3; //break; <---- commented out. And to every novices surprise execution just carries on from the 'matching' case unless a break (or return) is encountered.
I'm sure it seemed like a good idea at the time and there are some really fruiting coding tricks that exploit it. Duff's device is the most famous.
http://en.wikipedia.org/wiki/Duff's_device
999 times out of a 1000 it's an annoying bug that will take an hour to find.