reference vs. pointer

[al.cpwn]

I get that these two are different

int* get()
{
static int m;
return &m;
}

int& get()
{
static int m;
return m;
}

int& get()
{
static int m;
int* p=&m;
return p; // should it be *p or p?
}
but can there be a case when pointers may be confused with references?
The reason I am asking is because the pointer contains the address of
an object so its value is a reference.

[Gavin Deane]

al*****@gmail.c om wrote:

Tomás wrote:

int& = *new int;

Typo:
int& i = *new int;
-Tomás

Interesting. Let me make an attempt to see what the code does.
i is of type int&
*new int = *(new int) = *(pointer to memory allocated to store an
integer) = **(memory allocated for int)

thus if i=*new int; then i should be of type **?

No, you've gone the wrong way in your last step.

*new int => *(new int)
*(new int) => *(pointer to an int)

Now what happens when you apply * the dereference operator to a
pointer? You get the object pointed to by that pointer, so

*(poiner to an int) => (the int)

So in the code

int& i = *new int;

i is a reference to the newly created int.

Gavin Deane

[al.cpwn]

Gavin Deane wrote:

al*****@gmail.c om wrote:

Tomás wrote:

> int& = *new int;

Typo:
int& i = *new int;
-Tomás

Interesting. Let me make an attempt to see what the code does.
i is of type int&
*new int = *(new int) = *(pointer to memory allocated to store an
integer) = **(memory allocated for int)

thus if i=*new int; then i should be of type **?

No, you've gone the wrong way in your last step.

*new int => *(new int)
*(new int) => *(pointer to an int)

Now what happens when you apply * the dereference operator to a
pointer? You get the object pointed to by that pointer, so

*(poiner to an int) => (the int)

So in the code

int& i = *new int;

i is a reference to the newly created int.

Gavin Deane

Ah yes, you are right. Blunder! And to think I'm going to work at a
financial company this summer!

[Jakob Bieling]

al*****@gmail.c om wrote:

Jakob Bieling wrote:

al*****@gmail.c om wrote:

in*****@gmail.c om wrote:
al*****@gmail.c om wrote:
> int& get()
> {
> static int m;
> int* p=&m;
> return p; // should it be *p or p?

It needs to be *p. Otherwise it will not compile.

thats true, but WHY?

Because they are different types.

Is there a difference between a C reference and a C++ reference? By C
reference I mean
int& a;
type declarations.

As far as I know, C does not have any references .. ?
--
jb

(reply address in rot13, unscramble first)

[Tomás]

No, you've gone the wrong way in your last step.

*new int => *(new int)
*(new int) => *(pointer to an int)

Now what happens when you apply * the dereference operator to a
pointer? You get the object pointed to by that pointer, so

*(poiner to an int) => (the int)

So in the code

int& i = *new int;

i is a reference to the newly created int.

Gavin Deane

Here's how I like to think of it.

int &i = *new int;

First of all, that's the same as:

int &i = * (new int);

The thing on the left hand side of the assignment is of the type "int". The
thing on the left hand side of the assignment must an l-value of type "int".
Let's break it down into expressions. First let's look at the following
expression:

new int

The type of this expression is "int *", as "new" returns a pointer to the
object that's created. If you stick an asterisk behind an expression of type
"int*", you're left with an expression of type "int". Therefore, the
following expression is of the type "int":

*new int

Now we can bind a reference to it.

int &i = *new int;

delete &i;

-Tomás