Bit fields

[juice]

int main()
{
	struct personal
	{
		signed x : 2;
	}emp;
    emp.x=-3;
    printf("%d\n",emp.x);
}

the above code outputs 1 and the code below would output -1. Can someone please explain me the difference in output and how -3 gets stored in 2 bits. I am familiar with the concept of 2's complement.

int main()
{
	struct personal
	{
		signed x : 2;
	}emp;
    emp.x=3;
    printf("%d\n",emp.x);
}

[iMalc]

There are only four possible values that can be held in a signed two-bit number. What do you think those four values are?
Hint: -3 is certainly not one of them!

[nonoob]

In the first example you are assigning the last two bits of ...111101 (-3) to 2 bits. That means you end up with 01 = 1.
In the second example you are assigning the last two bits of ...0000011 (+3) to 2 bits. That means you end up with 11 = -1.

[juice]

In the first example you are assigning the last two bits of ...111101 (-3) to 2 bits. .

How did you get ....111101 for -3....
Isn't it supposed to be 111....111 for machines which copy the sign bit to all the vacant bits, and 100.....011 for machines which don't...

[juice]

Besides, we are only using 2 bits, out of which one is supposed to be sign bit..

[juice]

There are only four possible values that can be held in a signed two-bit number. What do you think those four values are?
Hint: -3 is certainly not one of them!

...are they 0, -1, +1 and -2...

[iMalc]

...are they 0, -1, +1 and -2...

Yep, well done.
You do indeed appear to be familiar enough with two's-complement.
I guess everything has been answered then.

[nonoob]

How did you get ....111101 for -3....
Isn't it supposed to be 111....111 for machines which copy the sign bit to all the vacant bits, and 100.....011 for machines which don't...

You need to review 2's complement.

...000000 = 0
...111111 = -1
...111110 = -2
...111101 = -3

That last two bits of -3 are 01.