for problem1 you can use %hi or %hd .. while scanning ..

On Thu, Jul 14, 2011 at 12:03 PM, Gaurav Jain <[email protected]>wrote:

> @Nicks
> *Problem 1*
>
> %d is used to take a signed integer as input. To take a short integer as
> input, use %hi. That way, you would get the correct answer as 2.
>
> *Problem 2:*
> a:1 means that variable a is of width *1 bit*
> Similarly, b:2 means that b is of width *2 bits*
>
> b = 6 sets the two bits as 10, (last two bits of 110 considered), which is
> equal to -2
> a = 2 sets the only bit as 0, (last bit of 10 considered), which is nothing
> but zero.
>
> Bit-fields like these however tend to be implementation-dependent and in
> the interest of portability should be avoided.
>
> t.b = 6 sets the last two bits of b as 10, which is -2 in 2's complement
> t.a = 2 sets the
> On Thu, Jul 14, 2011 at 1:18 AM, nicks <[email protected]> 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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>
>
>
> --
> Gaurav Jain
> Associate Software Engineer
> VxVM Escalations
> Symantec Software India Pvt. Ltd.
>
>
>
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