Struct/Union pointer stuff

[Kenneth Bull]

Say,

struct foo
{
int x;
double y;
/* etc. more variables defined */
short u;
char z;
} a;


Are the following true on all systems?

1) &a == &a.x
2) (&a.x < &a.y) && ( .. etc.. ) && ( &a.u < & a.z)


Are the following true if we change 'struct' to 'union' above, on all
systems?

3) &a == &a.x
4) &a.? == &a.? where ?s can be x, y, ..., u, or z (not necessarily
equal for both ?s)


Thank you for the help.

[CBFalconer]

Re: Struct/Union pointer stuff

Kenneth Bull wrote:

>
struct foo {
int x;
double y;
/* etc. more variables defined */
short u;
char z;
} a;
>
Are the following true on all systems?
>
1) &a == &a.x
2) (&a.x < &a.y) && ( .. etc.. ) && ( &a.u < & a.z)
>
Are the following true if we change 'struct' to 'union' above, on all
systems?
>
3) &a == &a.x
4) &a.? == &a.? where ?s can be x, y, ..., u, or z (not necessarily
equal for both ?s)

Yes to all.

--
[mail]: Chuck F (cbfalconer at maineline dot net)
[page]: <http://cbfalconer.home .att.net>
Try the download section.

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[Peter Nilsson]

Re: Struct/Union pointer stuff

Kenneth Bull wrote:

Say,
>
struct foo
{
int x;
double y;
/* etc. more variables defined */
short u;
char z;
} a;
>
>
Are the following true on all systems?
>
1) &a == &a.x
2) (&a.x < &a.y) && ( .. etc.. ) && ( &a.u < & a.z)

No, but you're almost there.

Note that the pointer types don't match. Your expressions
violate constraints. It is true to say that...

(int *) &a == &a.x
&a == (struct foo *) &a.x

offsetof(struct foo, x) == 0
offsetof(struct foo, x) < offsetof(struct foo, y)
offsetof(struct foo, y) < offsetof(struct foo, u)
offsetof(struct foo, u) < offsetof(struct foo, z)

Are the following true if we change 'struct' to 'union' above, on all
systems?
>
3) &a == &a.x

(int *) &a == &a.x
&a == (union foo *) &a.x

4) &a.? == &a.? where ?s can be x, y, ..., u, or z (not necessarily
equal for both ?s)

If the addresses are converted to void * or suitable type, they must
compare equal.

Note that it is not possible to take the address of bitfields.

--
Peter