Re: Writing Efficient Arguments

From:

gpuchtel <gpuchtel@gmail.com>

Newsgroups:

comp.lang.c++

Date:

Tue, 10 Jul 2007 08:16:59 -0700

Message-ID:

<1184080619.773537.269460@d55g2000hsg.googlegroups.com>

On Jul 10, 8:13 am, Michael DOUBEZ <michael.dou...@free.fr> wrote:

gpuchtel a =E9crit :

On Jul 10, 6:41 am, Michael DOUBEZ <michael.dou...@free.fr> wrote:

AFAIK the following code is valid (though UB):

void print(int& i)
{
  cout<<&i<<endl;

}

int main()
{
     print(*static_cast<int*>(0));

     return 0;

}

And prints '0'.

A reference cannot be 'null'. What you did created a tempory int
variable that was initialized to 0 (zero), then a reference to that
temporary variable was passed to the print function, which in turn
printed '0'.

Where did you get that from ? Even if I wanted it, I couldn't pass a
temporary because the ref is not const.

It is perhaps more understandable if you replace by:
int *ptr=NULL;
print(*ptr)

and add in print
++i; //cause segfault;

Michael- Hide quoted text -

- Show quoted text -

I didn't say (or mean) 'you' would pass a temporary, I was implying
what the compiler will do (pass). Simply stated, there is no such
thing as a 'null' reference. All your example does is to create a
reference to a (tempory) int value that contains the value of zero.
Your example of a null pointer is not relevant to the discussion of
the (im)possibility of having a 'null' reference.