difference between *foo and (*foo) ?

[Manuel]

Hi,

I've read that a line like this:

square *s1 = new square(0,0,20,20);

- alloc the memory needed for obj square,
- create a new object ,
- and return the address into the pointer s1

so, if I use the dereference operator, I access the obj.
The square obj has a method called "setBackgroundColor", so I thinked
to use simply:

*s2.setBackgroundColor(0.0,0.0,0.8,0.5);

but this not work:
- request for member of non-aggregate type before '(' token

Instead, if I use:

(*s2).setBackgroundColor(0.0,0.0,0.8,0.5);

it work.

I think that the first don't work because the compiler consider

*s2.setBackgroundColor(0.0,0.0,0.8,0.5);
as
*(s2.setBackgroundColor(0.0,0.0,0.8,0.5));

but it's only a my idea, so I'm looking for a "real" explanation :-)

thx,

Manuel

[Neelesh Bodas]

Manuel wrote:

I think that the first don't work because the compiler consider

*s2.setBackgroundColor(0.0,0.0,0.8,0.5);
as
*(s2.setBackgroundColor(0.0,0.0,0.8,0.5));

but it's only a my idea, so I'm looking for a "real" explanation :-)

The member selection operator (.) has a higher precedence than the
dereference operator.

[Russell Wood]

On 2006-01-06, Manuel <ma**********************@tin.it> wrote:

Hi,

I've read that a line like this:

square *s1 = new square(0,0,20,20);
...
so, if I use the dereference operator, I access the obj.
The square obj has a method called "setBackgroundColor", so I thinked
to use simply:

*s2.setBackgroundColor(0.0,0.0,0.8,0.5);

but this not work:
- request for member of non-aggregate type before '(' token

Instead, if I use:

(*s2).setBackgroundColor(0.0,0.0,0.8,0.5);

it work.

It's because your object is a pointer:

s2->setBackgroundColor(...);

- Russell

[Victor Bazarov]

Manuel wrote:

I've read that a line like this:

square *s1 = new square(0,0,20,20);

- alloc the memory needed for obj square,
- create a new object ,
- and return the address into the pointer s1
Rather, "- and initialise the pointer 's1' with that address".
so, if I use the dereference operator, I access the obj.
Yes.
The square obj has a method called "setBackgroundColor", so I thinked
to use simply:

*s2.setBackgroundColor(0.0,0.0,0.8,0.5);

but this not work:
- request for member of non-aggregate type before '(' token

Instead, if I use:

(*s2).setBackgroundColor(0.0,0.0,0.8,0.5);

it work.
"Dot" operator has precedence over the "asterisk" operator.
I think that the first don't work because the compiler consider

*s2.setBackgroundColor(0.0,0.0,0.8,0.5);
as
*(s2.setBackgroundColor(0.0,0.0,0.8,0.5));

but it's only a my idea, so I'm looking for a "real" explanation :-)

Find a decent book where operators and their precedence are explained.

V

[Manuel]

Russell Wood wrote:

s2->setBackgroundColor(...);

ok.

So

s2->setBackgroundColor(...);

is perfectly equal to

(*s2).setBackgroundColor(...);
?

[Victor Bazarov]

Manuel wrote:

Russell Wood wrote:

s2->setBackgroundColor(...);

ok.

So

s2->setBackgroundColor(...);

is perfectly equal to

(*s2).setBackgroundColor(...);
?

In your case, yes. IOW, if 's2' is a pointer, then they are the same.

If 's2' is an object, you can have operator-> overloaded for it and
operator* overloaded, in such a way that those two expressions do not
produce the same result. But I am guessing that you're not at that
stage yet.

V

[Russell Wood]

On 2006-01-06, Manuel <ma**********************@tin.it> wrote:

Russell Wood wrote:

s2->setBackgroundColor(...);

ok.

So

s2->setBackgroundColor(...);

is perfectly equal to

(*s2).setBackgroundColor(...);

Correct.