struct and members addresses

struct and members addresses

[Cédric Lucantis]

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Hello

   I would like to know if I can assume that a structure and its first member
will always have the same address. For instance, with the following code:

/* ----------------------------------*/

struct foo
{
   any_type    first_member;
};

struct foo my_struct;

void * addr1 = &my_struct;
void * addr2 = &my_struct.first_member;

/* ----------------------------------*/

can I be sure that `addr1' will be equal to `addr2' ?

thanks


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Re: struct and members addresses

[Alexandre Courbot]

> struct foo
> {
>    any_type    first_member;
> };
>
> struct foo my_struct;
>
> void * addr1 = &my_struct;
> void * addr2 = &my_struct.first_member;
>
> /* ----------------------------------*/
>
> can I be sure that `addr1' will be equal to `addr2' ?

In C, I think you can safely. In C++, if your struct has virtual methods, the 
adress won't be the same since I think GCC puts the vtable at the beginning 
of the struct. But as long as you don't use virtuals, it should be ok AFAIK.

Alex.

Re: struct and members addresses

[Andrea 'fwyzard' Bocci]

At 13.00 24/04/2003 +0100, Cédric Lucantis wrote:
>Hello
>
>    I would like to know if I can assume that a structure and its first member
>will always have the same address. For instance, with the following code:
>
>/* ----------------------------------*/
>
>struct foo
>{
>    any_type    first_member;
>};
>
>struct foo my_struct;
>
>void * addr1 = &my_struct;
>void * addr2 = &my_struct.first_member;
>
>/* ----------------------------------*/
>
>can I be sure that `addr1' will be equal to `addr2' ?
>
>thanks

I think so.
IF yours is a simple struct, ie. no hereditaritey or any C++ thing.

You can check that offsetof (my_struct, first_member) is 0.

.fwyzard. 

RE: struct and members addresses

[Agust Karlsson]

Or to be sure you could use
__attribute__ ((packed)) 

I.e.
	 struct foo
	 {
	     any_type    first_member;
	 } __attribute__ ((packed)) ;

	Gústi
--
Agust Karlsson            mailto:gusti@pallas.dk
Pallas Informatik A/S     http://www.pallas.dk
Allerød Stationsvej 2D    Tel.: +45 48 10 24 10
DK-3450 Allerød           Fax.: +45 48 10 24 01


> -----Original Message-----
> From:	Alexandre Courbot [SMTP:Alexandre.Courbot@lifl.fr]
> Sent:	Thursday, April 24, 2003 1:16 PM
> To:	Cédric Lucantis; gcc-help@gcc.gnu.org
> Subject:	Re: struct and members addresses
> 
> 
> > struct foo
> > {
> >    any_type    first_member;
> > };
> >
> > struct foo my_struct;
> >
> > void * addr1 = &my_struct;
> > void * addr2 = &my_struct.first_member;
> >
> > /* ----------------------------------*/
> >
> > can I be sure that `addr1' will be equal to `addr2' ?
> 
> In C, I think you can safely. In C++, if your struct has virtual methods, the 
> adress won't be the same since I think GCC puts the vtable at the beginning 
> of the struct. But as long as you don't use virtuals, it should be ok AFAIK.
> 
> Alex.
> 

RE: struct and members addresses

[Martin York]

Just to throw a spanner in the works!
Assuming C only

What happens if there are more members of the structure.
Is the compiler allowed to reorder the members of the
structure for efficiency in memory layout?




-----Original Message-----
From: Alexandre Courbot [mailto:Alexandre.Courbot@lifl.fr]
Sent: 24 April 2003 07:16
To: Cédric Lucantis; gcc-help@gcc.gnu.org
Subject: Re: struct and members addresses



> struct foo
> {
>    any_type    first_member;
> };
>
> struct foo my_struct;
>
> void * addr1 = &my_struct;
> void * addr2 = &my_struct.first_member;
>
> /* ----------------------------------*/
>
> can I be sure that `addr1' will be equal to `addr2' ?

In C, I think you can safely. In C++, if your struct has virtual methods,
the 
adress won't be the same since I think GCC puts the vtable at the beginning 
of the struct. But as long as you don't use virtuals, it should be ok AFAIK.

Alex.

RE: struct and members addresses

[Rupert Wood]

Martin York wrote: 

> Just to throw a spanner in the works!
> Assuming C only
>
> What happens if there are more members of the structure.
> Is the compiler allowed to reorder the members of the structure for
> efficiency in memory layout?

No. C99 6.2.5.20 second bullet - "sequentially allocated":

    A /structure type/ describes a sequentially allocated nonempty set
    of member objects (and, in certain circumstances, an incomplete
    array), each of which has an optionally specified name and
    possibly distinct type.

Minor exception for adjacent bit-fields combined into a single unit; 6.2.7.1
bullets 8 and 10:

    A member of a structure or union may have any object type other
    than a variably modified type. In addition, a member may be
    declared to consist of a specified number of bits (including a
    sign bit, if any). Such a member is called a /bit-field/; its
    width is preceded by a colon.

    An implementation may allocate any addressable storage unit large
    enough to hold a bit-field. If enough space remains, a bit-field
    that immediately follows another bit-field in a structure shall be
    packed into adjacent bits of the same unit. ... The order of
    allocation of bit-fields with a unit (high-order to low-order or
    low-order to high-order) is implementation defined. The alignment
    of the addressable storage unit is unspecified.