Struct - Explain the output:

Question

I have the following C code.

struct values{
  int a:3;
  int b:3;
  int c:2;
};

void main(){
  struct values v={2,-6,5};
  printf(\"%d %d %d\",v.a,v.b,v.c);         


        

Answer 1

{2,             |  -6,             |        5        }
010 last 3 bits |  010 last 3 bits |  01 last 2 bits
2                  2                        1    

Answer 2

-6 exceeds the range of a 3-bit signed int. Therefore you're observing an artifact of undefined implementation-defined behaviour (in practice, the most-significant bits of your value are being thrown away).

Answer 3

No. Output is 2 2 1.

The C compiler converts the values to Binary, and stores in the memory.

Binary value of 2 : 00000010

Binary value of -6: 11111010 (11111001+1)

Binary value of 5 : 00000101

While storing in memory:

For 2, 010 will be stored.

For -6, 010 will be stored.

For 5, 01 will be stored.

When you access these variables from your main method, for v.a "010" will be returned, here left most bit is for sign.

So v.a is 2. Similarly v.b is 2 and v.c is 1.

Hope it helps.