#include<stdio.h>
main()
{
struct value
{
int bit1 : 1;
int bit2 : 4;
int bit3 : 4;
}bit={1, 2, 2};
printf("%d %d %d\n",bit.bit1,bit.bit2,bit.bit3);
}
Output of this code is "-1 2 2" Please clarify the logic behind this output. Value of bit.bit2 and bit.bit3 is always same as the value assigned to it but bit.bit1 is changing with different integer values. why?
You should use unsigned int. The highest bit, defines whether a number is negative or positive if signed values are used. If you have only one bit, and this is 1, then it it is interpreted as a negative number as the highest bit is set.
If you set the other values to 15 you will also get a negative output.
You could modify the output by using %u in the printf command, but you would still have possibly unwanted effects when assigning and comparing it with other values.
int x : b ; means you are allocating only b bits of memory to x instead of the default sizeof(int) bytes. This kind of declaration is only possible inside a structure.
Range of signed integer in C is -2^(b-1) to 2^(b-1)-1. Where b is number of bits used to store the integer. In all the above cases overflow occurs. A good compiler should give you a warning about overflow.
A signed bit-field of size 1 accepts values in the range [-1 … 0]. This is a consequence of the general formula [-2^(N-1) … 2^(N-1)-1] for determining the range of values that can be stored in N bits with 2's complement representation for N=1.
If you expected bit1 to hold the values 0 or 1, you can declare it as unsigned int bit1 : 1;.
The standard does not specify whether int in bit-fields is signed or unsigned. Instead, it forces you to explicitly specify signedness.
A bit-field shall have a type that is a qualified or unqualified version of
_Bool,signed int,unsigned int, or some other implementation-defined type.
