Re: Pointer or reference?

From:

SG <s.gesemann@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Wed, 14 Oct 2009 06:31:00 -0700 (PDT)

Message-ID:

<b05b4114-a215-4dfd-8bb8-9dcf763bd21b@c3g2000yqd.googlegroups.com>

On 14 Okt., 14:33, davee <da...@nomail.com> wrote:

Lets assume that I have an object 'MyObject' with various fields. Below
foo1 and foo2 modifies the fields in a MyObject object:

void foo1(MyObject * mobj){
   int value = 55;
   myobj->setValue(value);
   ...
   ...

}

void foo2(MyObject *& mobj){
   int value = 55;
   myobj->setValue(value);
   ...
   ...

}

int main(){
   MyObject * myobj = new MyObject();
   foo1(myobj);
   std::cout << "getValue = " << myobj.getValue() << std::endl;

   foo2(myobj);
   std::cout << "getValue = " << myobj.getValue() << std::endl;
}

For the record: MyObject is not an object but a type, specifically a
class-type.

As I understand its only when calling foo2 that the changes will visible
in main after returning from the call. The reason is that foo1 works
on a copy of the pointer which disappears when the function returns.
On contrary foo2 works on the original pointer. Or am I mistaken ?

Have you tried?

You don't seem to differentiate between the pointer and the pointee.
In the first version you pass a pointer by value, in the second by
reference. But "mobj" which in in both cases refers to a pointer to a
MyObject object is not modified so it does not matter which function
you call. Both functions will however mutate the object pointed to --
visibly. So, the expected output is

getValue = 55 (as set in foo1)
getValue = 55 (as set in foo2)

See "pointer fun with Binky" (C++ version)
http://www.youtube.com/watch?v=i49_SNt4yfk

Cheers,
SG