@gaurav :y it is -2?y not +2?

On Sat, Jul 16, 2011 at 2:13 PM, sukhmeet singh <[email protected]>wrote:

> 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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-- 
Regards,
Kamakshi
[email protected]

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