CODEWITHSUNDEEP
cstruct
Raja Narayan
Raja Narayan

Reputation: 595

Using bit with int in structure

   #include<stdio.h>
   struct a
   {
      int a:4;
   };
   main(){
   struct a aa;
   aa.a=9;
   printf("a=%d\n",aa.a);
   return 0;
   }

Here the output is -7. Why is it so? what does exactly int a:4 does ? please explain

Upvotes: 6

Views: 138

Answers (4)

Steve P.
Steve P.

Reputation: 14699

Since it's two's complement, the highest order bit is used for the sign. By writing a:4 you're saying to only allocate 4 bits of memory, which leaves 3 bits left over for the actual number. So our effective range is [-8,7]. Since all 1's is -1, there's an extra number on the negative side. For more of an explanation on this, see the above link.

9, in (unsigned) binary is: 1001. When you put this into a (signed), you get that a is negative, due to the initial 1, and since the following numbers are 001, we add 1 to the max negative number, thereby giving us -7.

If you want to store the number 9 in only 4 bits, you need to use an unsigned int, which would give you a range of [0, 15].

EDIT:
In case anyone is struggling with figuring out how 1001 signed gives us -7, consider the following:

Since 1111 is -1, let some variable value = -1.

To figure out the values of a negative (signed) int num, let us denote xi in num:

xi : {0,1 at position i, where i=0 is the least significant bit)},

Then, for every xi = 0, subtract 2i from value.

Example:

1001:

value = -1 - 21 - 22 = -7

Upvotes: 5

user2107435
user2107435

Reputation:

you have to use unsigned indeed int :

#include<stdio.h>
struct a
{
  unsigned a:4; //  <-- unsigned indeed int, then work good
};
main(){
struct a aa;
aa.a=9;
printf("a=%d\n",aa.a);
return 0;
}

output :

   a=9

Upvotes: 2

shf301
shf301

Reputation: 31394

Your field is a 4 bit signed integer. For signed integers the upper bit is a sign bit, which means that you only have 3 bits for the actual number. The range of numbers you can store in the field are -8 to 7 (assuming 2's compliment storage).

The bit pattern for 9 is 1001, which has the 4th bit set, meaning it is interpreted as a negative number, which is why it is printing out as a -7. If you would have expected a -1, you need to read up on 2's compliment.

If you want to be able to store 9 in the field, make a an unsigned int

Upvotes: 4

Jens Gustedt
Jens Gustedt

Reputation: 78903

You only reserved 4 bits for the field, one bit is used for the sign, so only 3 bits remain for positive values. Thus you can only store values up to 7.

Upvotes: 3

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