C Snippet behavior having struct bitfield of unsigned variables

Question

Please help me out with logic associated with code snippet below. The output I get is 12 8. How come *p is 12? I understood its size to be 8, but I'm perturbed with value displayed for *p.

int main()    
{    
    struct bitfield
    {
        unsigned a:5;
        unsigned c:5;
        unsigned b:6;
    } bit;

    char *p;

    struct bitfield bit1={1,3,3};

    p=(char*)&bit1;

    p++;

    printf("%d\t%d",*p,sizeof(bit1));

    getchar();

    return 0;
}

Note that almost everything to do with bit-fields w.r.t memory layout and the like is implementation defined. Accessing the bit-field values via the char * is implementation defined at best. It would certainly be feasible for the value displayed to be all zero on some types of processor. — Jonathan Leffler
– Jonathan Leffler, Commented Dec 19, 2015 at 17:20

Ian · Accepted Answer · 2015-12-19 17:29:53Z

3

Firstly,

because of the declaration

struct bitfield
{
    unsigned a:5;
    unsigned c:5;
    unsigned b:6;
}bit;

Your bit-field bit will have 16-bit bitfield with the arrangement like this

00000000 00000000
ccccccbb bbbaaaaa

Secondly, because the input is {1, 3, 3}, so the 1 will be assigned to a, 3 to b, and another 3 to c, makes it become like this

00001100 01100001
ccccccbb bbbaaaaa

Thirdly, because the p is just character pointer, it will originally take the last byte (an 8-bit) among the two, which has smaller address in little endian format (edit: albeit implementation/platform dependent, see comment and another answer from Abstraction), so, p points to

Additional note from MDSN

The 8086 family of processors stores the low byte of integer values before the high byte

01100001
bbbaaaaa

But fourthly, you increase the p! Thus, in little endian format, it now points to

00001100
ccccccbb

Then finally, you print the value of what is pointed by p. And obviously, 00001100 is nothing but bit representation of decimal value 12. That is why you get 12.

edited Dec 19, 2015 at 17:29

answered Dec 19, 2015 at 16:33

Ian

30.9k20 gold badges76 silver badges112 bronze badges

Sign up to request clarification or add additional context in comments.

3 Comments

honeytechiebee Over a year ago

nicely explained. Thanks Ian.

Ian Over a year ago

No problem, I am glad that you find it useful. It once confused me too. =)

Jonathan Leffler Over a year ago

Note that this layout is, like almost everything to do with bit-fields, implementation specific. The layout, packing, etc of the bits depends on the implementation.

Abstraction · Accepted Answer · 2015-12-19 16:49:19Z

3

This is very platform/implementation dependent code. Assuming that the values are tightly packed in 16bit integer and the endianness is little endian then your values should look like (left -> right | lower memory address -> higher memory address)

   1     3     3
  ||     ||    ||
10000 110 00 110000
     ^   ^  ^
     5   8  10

Now 00110000 is 12 (the notation is from left to right least to most significant and 00001100 in the standard notation). And your p points after the first 8 bits since you increment it.

edited Dec 19, 2015 at 16:49

answered Dec 19, 2015 at 16:43

Abstraction

5112 silver badges9 bronze badges

Collectives™ on Stack Overflow

C Snippet behavior having struct bitfield of unsigned variables

2 Answers 2

3 Comments

Comments

Your Answer

Hot Network Questions

Collectives™ on Stack Overflow

2 Answers 2

3 Comments

Comments

Your Answer

Sign up or log in

Post as a guest

Related