your program enter at
case 3:
after that it kept falling in next cases as well.... hence x was assigned
values
x=0,
x=0,
and x=4 at the end.. and then exit the switch case...
use print after every assignment u'll see that...
On Fri, Jun 29, 2012 at 3:16 PM, saurabh singh <saurab...@gmail.com> wrote:
> 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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