Using bit with int in structure

Question

   #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

Steve P. · Accepted Answer · 2013-10-09 05:43:06Z

5

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 x_i in num:

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

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

Example:

1001:

value = -1 - 2¹ - 2² = -7

edited Oct 9, 2013 at 5:43

answered Jul 13, 2013 at 17:40

Steve P.

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4 Comments

Carl Norum Over a year ago

The int in a bitfield might be unsigned int by default. It appears not to be in the OP's case, but it never hurts to read the documentation.

lsk Over a year ago

Nice question and very nice answer! I verified that the unsigned type will give the expected answer.

Doug Currie Over a year ago

I prefer to think of the MSB as having the value -2^3 + 2^0 -- that way all the non-MSBs have their usual value.

Steve P. Over a year ago

Yeah, that's how I was thinking about it above the edit, but I wanted to show an alternative too.

shf301 · Accepted Answer · 2013-07-13 17:38:21Z

4

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

answered Jul 13, 2013 at 17:38

shf301

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Comments

Jens Gustedt · Accepted Answer · 2013-07-13 17:34:36Z

3

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.

answered Jul 13, 2013 at 17:34

Jens Gustedt

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1 Comment

Raja Narayan Over a year ago

could u please explain using binary representation?

user2107435 · Accepted Answer · 2013-07-13 17:42:11Z

2

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

answered Jul 13, 2013 at 17:42

user2107435

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