Is this notation correct

[johnnash]

typedef astruct
{
int x;

} A;

struct b

{
A a;

}B;

struct b *C;

C = &B;

so if i want to access say B.a.x, and if i am going to use the
pointer notation(using C), then is it right to say -

1. (C->a).x

2. (*C).a.x

[johnnash]

Re: Is this notation correct

im sorry if i am bugging any of you with my silly questions but the
thing is im just a intermediate level programmer and my i haven't
touched C for almost 4 years so it would be nice if some one could
answer this one and also suggest to me some nice tutorial for revising
key concepts like the one used in this problem.

[Eric Sosman]

Re: Is this notation correct

johnnash wrote:

typedef astruct
{
int x;
>
} A;
>
struct b
>
{
A a;
>
}B;
>
struct b *C;
>
C = &B;
>
so if i want to access say B.a.x, and if i am going to use the
pointer notation(using C), then is it right to say -
>
1. (C->a).x
>
2. (*C).a.x

Both are right. `C->a.x' is also right, and is easier
to type.

--
Eric.Sosman@sun .com

[Richard Heathfield]

Re: Is this notation correct

johnnash said:

typedef astruct
{
int x;
} A;
struct b
{
A a;
}B;
struct b *C;
C = &B;
>
so if i want to access say B.a.x, and if i am going to use the
pointer notation(using C), then is it right to say -
>
1. (C->a).x

Yes, but C->a.x will be fine too. You don't need the parentheses in this
situation.

2. (*C).a.x

That works too. They are synonymous.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999

[johnnash]

Re: Is this notation correct

>

2. (*C).a.x

>
That works too. They are synonymous.
>

Would it be correct to remove the bracket in this case as well..i mean

*C.a.x

as B = *C

[Richard Heathfield]

Re: Is this notation correct

johnnash said:

>

>>

2. (*C).a.x

>>
>That works too. They are synonymous.
>>

>
>
Would it be correct to remove the bracket in this case as well..i mean
>
*C.a.x
>
as B = *C

No. That's because . has a higher precedence than *

Precedence is a way of thinking about how operands bind to operators.
Because . has a higher precedence than *, *C.a.x is equivalent to *(C.a.x)
- that is, if x were a pointer, you'd be dereferencing it (and because it
isn't a pointer, the code is illegal). So, to insist that the * belongs to
the C object rather than C.a.x, we have to put parentheses around it.

Alternatively, you can use C->a.x, which is much clearer anyway.

--
Richard Heathfield <http://www.cpax.org.uk >
Email: -http://www. +rjh@
Google users: <http://www.cpax.org.uk/prg/writings/googly.php>
"Usenet is a strange place" - dmr 29 July 1999