Any help please

Question

I don't understand why it's not printing case 3.

#include <stdio.h>
main(){
int number=0;
while(number<5) {
switch (number) {
default:
printf("defualt\n");
case 0 :
printf("cas 0\n");
break;
case 3 :
printf("cas 3\n");
break;
case 2 :
++number;
case -1 :
printf("cas -1\n");
break;
case 1 :
printf("cas 1\n");
}
++number;
}
return 0;
}

Answer 1

case 3 is not printed because in case 2 number is incremented to 3, but after the switch statement number is incremented again (to 4)

Answer 2

Please don't just post code - communicate!

What is your code supposed to do?

What happens instead?

What have you tried to fix it?

How can we help you?

Answer 3

what is wrong with that code?

a function must have a return type (void if it doesn't return something)
you returned 0 at the end of the code, so you should also add the return type (int) to de function definition

int main () {
...

Answer 4

Mention complete detail what output you want through this code and what i have to fix in this code and choose relevant tag according to your problem.

Answer 5

The only thing I can see is "wrong/missing" is a "break" in case 2 so it falls though to case -1......but without knowing what you'r trying to do...It's had to say.
edit..and a "break" at the default as it's not at the end of the switch.

Answer 6

Just I found in axm
I don't understand why not printing case 3

Answer 7

What should be the output

Answer 8

You'r incrementing at case 2 as well as at the end.

Answer 9

@John Robotane
that's probably just missing when the code was mis-copied and pasted onto here.

Answer 10

ya