the cases are simple lables they have nothing to do with the flow of
program.
Saurabh Singh
B.Tech (Computer Science)
MNNIT
blog:geekinessthecoolway.blogspot.com
On Fri, Jun 29, 2012 at 3:14 PM, adarsh kumar <algog...@gmail.com> wrote:
> Doubt, very trivial though:
> #include<stdio.h>
> int main()
> {
> int x=3;
> switch(x)
> {
> case 1:
> x=1;
> break;
> case 2:
> x=2;
> break;
> case 3:
> x=3;
> break;
> default:
> x=0;
> break;
> case 4:
> x=4;
> break;
> }
> printf("%d",x)
> return 0;
> }
> gives an output of 3. But,
> #include<stdio.h>
> using namespace std;
> int main()
> {
> int x=3;
> switch(x)
> {
> case 1:
> x=1;
> case 2:
> x=2;
> case 3:
> x=3;
> default:
> x=0;
> case 4:
> x=4;
> }
> printf("%d",x);
> getch();
> return 0;
> }
> gives an output of 4.
> My doubt is, in spite of the missing break statements in the second case,
> how will it enter case 4, as it should check if x=4 before doing that,
> which is not true.
>
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