Questions on bitfields and struct

Discussion in 'C Programming' started by sarathy, Jul 20, 2006.

  1. sarathy

    sarathy Guest

    Hi,

    1. How it that the results for the size of struct1 and struct2 (below)
    are 4 and 3

    # include <stdio.h>

    struct struct1
    {
    const :16;
    volatile :4;
    };

    struct struct2
    {
    int :1;
    unsigned :1;
    const :16;
    volatile :4;
    };

    int main()
    {
    printf ("Size of struct 1 = %d\n",sizeof(struct struct1)); /*
    Prints 4 */
    printf ("Size of struct 2 = %d\n",sizeof(struct struct2)); /*
    Prints 3 */

    }

    2. Also what is meant by "incomplete type" in C?
    3. What does tag refer to in a struct/union declaration?

    Regards,
    Sarathy
    sarathy, Jul 20, 2006
    #1
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  2. In article <>,
    sarathy <> wrote:

    >1. How it that the results for the size of struct1 and struct2 (below)
    >are 4 and 3


    >struct struct1
    >{
    > const :16;
    > volatile :4;
    >};


    >struct struct2
    >{
    > int :1;
    > unsigned :1;
    > const :16;
    > volatile :4;
    >};


    With SGI's IRIX cc compiler, both are 3. With gcc 3.3 on the same
    machine, the first is 4 and the second is 3. Which is to say that the
    alignment mechanisms are implementation dependant, and are not required
    to be consistant.

    It is, for example, entirely legal for the compiler to examine the
    first structure, note that the first field is the same size as a short
    int on that implementation, and decide that the structure shall be
    built for fast access as a pair of shorts aligned on a short boundary,
    for a total of 4 bytes.

    The same compiler could look at the second structure and decide that it
    is complicated enough that space is the important factor rather than
    speed, and decide to pull bits out of 3 bytes.

    But compilers are not required to allow bitfields to cross word
    boundaries, so the compiler -could- have decided to put the first two
    fields into one short, the third field into a second short, and the
    fourth field into a third short, for a total of 6 bytes. Or it could
    have decided that since the total fits within 32 bits that it would
    pull bits out of a long, for a total of 4 bytes.

    Decisions about when to move to the next word and how much padding
    to use before that word are completely up to the compiler: the
    closest that the C standard comes on this point is to say "if it fits".


    >2. Also what is meant by "incomplete type" in C?


    The C faq probably talks about that in better words than I could
    come up with in a reasonable time.


    >3. What does tag refer to in a struct/union declaration?


    A structure tag is a programmer-given name for that variety of
    structure. It is not a variable, but rather a reference to the type.
    As a rough analogy: "Form 37/J" might be a name given to a particular
    layout of income tax form, but "Form 37/J" is not a particular -copy-
    of the form, it is the name of the -kind- of form.
    --
    I was very young in those days, but I was also rather dim.
    -- Christopher Priest
    Walter Roberson, Jul 20, 2006
    #2
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  3. sarathy

    sarathy Guest

    Walter Roberson wrote:
    > In article <>,
    > sarathy <> wrote:
    >
    > >1. How it that the results for the size of struct1 and struct2 (below)
    > >are 4 and 3

    >
    > >struct struct1
    > >{
    > > const :16;
    > > volatile :4;
    > >};

    >
    > >struct struct2
    > >{
    > > int :1;
    > > unsigned :1;
    > > const :16;
    > > volatile :4;
    > >};

    >
    > With SGI's IRIX cc compiler, both are 3. With gcc 3.3 on the same
    > machine, the first is 4 and the second is 3. Which is to say that the
    > alignment mechanisms are implementation dependant, and are not required
    > to be consistant.
    >
    > It is, for example, entirely legal for the compiler to examine the
    > first structure, note that the first field is the same size as a short
    > int on that implementation, and decide that the structure shall be
    > built for fast access as a pair of shorts aligned on a short boundary,
    > for a total of 4 bytes.
    >
    > The same compiler could look at the second structure and decide that it
    > is complicated enough that space is the important factor rather than
    > speed, and decide to pull bits out of 3 bytes.
    >
    > But compilers are not required to allow bitfields to cross word
    > boundaries, so the compiler -could- have decided to put the first two
    > fields into one short, the third field into a second short, and the
    > fourth field into a third short, for a total of 6 bytes. Or it could
    > have decided that since the total fits within 32 bits that it would
    > pull bits out of a long, for a total of 4 bytes.
    >
    > Decisions about when to move to the next word and how much padding
    > to use before that word are completely up to the compiler: the
    > closest that the C standard comes on this point is to say "if it fits".
    >
    >
    > >2. Also what is meant by "incomplete type" in C?

    >
    > The C faq probably talks about that in better words than I could
    > come up with in a reasonable time.
    >
    >
    > >3. What does tag refer to in a struct/union declaration?

    >
    > A structure tag is a programmer-given name for that variety of
    > structure. It is not a variable, but rather a reference to the type.
    > As a rough analogy: "Form 37/J" might be a name given to a particular
    > layout of income tax form, but "Form 37/J" is not a particular -copy-
    > of the form, it is the name of the -kind- of form.
    > --
    > I was very young in those days, but I was also rather dim.
    > -- Christopher Priest


    Thanks a lot. I got the things clear now. The next fun starts...
    Consider the code

    struct struct1
    {
    short a1:16;
    short b1:4;
    };



    struct struct2
    {
    int a2:16;
    unsigned b2:1;
    short c2 :16;
    short d2 :4;
    };



    int main()
    {
    struct struct1 *str1=(struct struct1 *)malloc(sizeof(struct
    struct1));
    str1->a1=24;
    str1->b1=4;
    print_bits(str1,sizeof(str1));
    printf ("Size of struct1 = %d\n",sizeof(struct struct1));
    /* Prints 4 */
    printf ("Size of str1 = %d\n",sizeof(str1)); /* Prints 4
    */



    struct struct2 *str2=(struct struct2 *)malloc(sizeof(struct
    struct2));
    str2->a2=1;
    str2->b2=1;
    str2->c2=6;
    str2->d2=12;
    print_bits(str2,sizeof(struct struct2));
    printf ("Size of struct2 = %d\n",sizeof(struct struct2));
    /* Prints 8 */
    printf ("Size of str2 = %d\n",sizeof(str2)); /* Prints 4
    */
    }


    Output:
    ----------

    00000000 00000100 00000000 00011000
    Size of struct1 = 4
    Size of str1 = 4
    00000000 00001100 00000000 00000110 00000000 00000001 00000000 00000001
    Size of struct2 = 8
    Size of str2 = 4

    Is there any difference in sizeof (struct tag_name) and
    sizeof(struct_instance_identifier)
    In the third case, clearly from the results, 8 is correct due to
    padding. But why is 4 being printed in the last printf.
    sarathy, Jul 20, 2006
    #3
  4. In article <>,
    sarathy <> wrote:
    >
    >Walter Roberson wrote:


    [lots]

    Please trim down quotations to just the portion needed for discussion.


    >Consider the code


    > struct struct2 *str2=(struct struct2 *)malloc(sizeof(struct
    >struct2));


    > printf ("Size of str2 = %d\n",sizeof(str2)); /* Prints 4
    >*/


    >Size of str2 = 4



    >Is there any difference in sizeof (struct tag_name) and
    >sizeof(struct_instance_identifier)


    No.


    >In the third case, clearly from the results, 8 is correct due to
    >padding. But why is 4 being printed in the last printf.


    Because str2 is not an instance of struct struct2.
    str2 is a *pointer* to an instance of struct struct2.
    sizeof(str2) is printing out the size of the pointer.

    --
    Okay, buzzwords only. Two syllables, tops. -- Laurie Anderson
    Walter Roberson, Jul 20, 2006
    #4
  5. sarathy

    sarathy Guest

    Hi,
    Does the 4 imply that the machine is 32 bit using 32 bit
    addressing?
    Does this have any impact on wordlength of the machine, because

    in Turbo C in Windows(DOS environment) int --> 16 bit == address
    length
    in gcc in i386 Linux int --> 32 bits == address length

    Regards,
    Sarathy
    sarathy, Jul 22, 2006
    #5
  6. On 22 Jul 2006 00:18:47 -0700, "sarathy" <>
    wrote:

    >Hi,
    > Does the 4 imply that the machine is 32 bit using 32 bit
    >addressing?


    I give up - what 4?

    > Does this have any impact on wordlength of the machine, because


    And what this?
    >
    > in Turbo C in Windows(DOS environment) int --> 16 bit == address
    >length
    > in gcc in i386 Linux int --> 32 bits == address length


    We normally avoid issues regarding specific compilers and operating
    systems. There are groups where those subjects are topical.


    Remove del for email
    Barry Schwarz, Jul 22, 2006
    #6
  7. On 22 Jul 2006 00:18:47 -0700, in comp.lang.c , "sarathy"
    <> wrote:

    >Hi,
    > Does the 4 imply that the machine is 32 bit using 32 bit
    >addressing?


    Which 4?

    --
    Please quote enough of the previous message for context. Even google
    now makes this quite easy.
    Mark McIntyre, Jul 22, 2006
    #7
  8. On 2006-07-22, sarathy <> wrote:
    > Does the 4 imply that the machine is 32 bit using 32 bit addressing?


    What 4 are you talking about?

    > Does this have any impact on wordlength of the machine, because in
    > Turbo C in Windows(DOS environment) int --> 16 bit == address length


    C has no concept of "wordlength".

    > in gcc in i386 Linux int --> 32 bits == address length


    This is true to the best of my knowledge, but your other questions make
    me wonder how you arrived at that figure. Not by multiplying anything by
    8, right?

    --
    Andrew Poelstra <website down>
    My server is down; you can't mail
    me, nor can I post convieniently.
    Andrew Poelstra, Jul 22, 2006
    #8
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