1st problem
format specifier is %d whereas memory is allocated only for short int
memory allocated-2bytes. but being stored in 4bytes.
so a is overwritten by b.
In dis case where u give a=1, b=1, a becomes become 0.
n so it gives:0+1=1
in d scanf if u use %hd instead u ll get d right output!!
On Thu, Jul 14, 2011 at 8:01 AM, nicks <crazy.logic.k...@gmail.com> wrote:
> Hey Guys, plz help me in getting these 2 C output problems
>
> *PROBLEM 1>.*
> *
> *
> *#*include<stdio.h>
> int main()
> {
> short int a,b,c;
> scanf("%d%d",&a,&b);
> c=a+b;
> printf("%d",c);
> return 0;
> }
> INPUT-
> 1 1
>
> OUTPUT
> 1
>
> i am not getting why 1 is coming in the output.....what difference is
> using short making in the code ???
>
>
> *PROBLEM 2>.*
> *
> *
> *
> *
> #include<stdio.h>
> main()
> {
> struct
> {
> int a:1;
> int b:2;
> }t;
> t.b=6;
> t.a=2;
> printf("%d %d",t.a,t.b);
> }
>
> OUTPUT
> 0 -2
>
> What does the statement a:1 and b:1 mean and what are they doing.....i am
> seeing them first time ever...hence not able to get the output....if someone
> has any idea plz help !!
>
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