operator+

[Gaijinco]

I was perfectly sure that this code had to work:

#include <iostream>

int main()
{
std::cout << operator+(3,4);
return 0;
}

If I can overload the operator+() it means that there is a base
function defined by the language, right?

[Jacek Dziedzic]

Re: operator+

Gaijinco wrote:

I was perfectly sure that this code had to work:
>
#include <iostream>
>
int main()
{
std::cout << operator+(3,4);
return 0;
}
>
If I can overload the operator+() it means that there is a base
function defined by the language, right?

Note that you _can't_ overload operator+() for int.

- J.

[Alf P. Steinbach]

Re: operator+

* Gaijinco:

I was perfectly sure that this code had to work:
>
#include <iostream>
>
int main()
{
std::cout << operator+(3,4);
return 0;
}

Nope.

If I can overload the operator+() it means that there is a base
function defined by the language, right?

No.

You can overload operator+ for enum and class type arguments, and for
enum or class type reference arguments, but then there's no built-in
implementation. I.e. there's no "base" function you can base your
implementation on. Note for enums: in the case of enum arguments to +,
without a user-defined operator+ for that type, the arguments are
promoted to int or higher and the + operator for that integer type is used.

You can not overload operators for built-in types such as int, nor for
pointer types.

--
A: Because it messes up the order in which people normally read text.
Q: Why is it such a bad thing?
A: Top-posting.
Q: What is the most annoying thing on usenet and in e-mail?

[Ron Natalie]

Re: operator+

Jacek Dziedzic wrote:

Gaijinco wrote:

>I was perfectly sure that this code had to work:
>>
>#include <iostream>
>>
>int main()
>{
> std::cout << operator+(3,4);
> return 0;
>}
>>
>If I can overload the operator+() it means that there is a base
>function defined by the language, right?

>
Note that you _can't_ overload operator+() for int.
>

Further, you can't get at the builtin operator implementations
by trying to access them as overload functions. It's primarily
for this reason that funky little template functions like "less"
exist.

[Gaijinco]

Re: operator+

I always assumed that when I right in a code

a + b

What really happened is that the language rewrote the code to

operator+(a,b)

and that because of it, it was possible to use the function notation
for all operators.

So is there anyway in which I can use the function notation instead of
the classic notation?

[Nate Barney]

Re: operator+

Gaijinco wrote:

I always assumed that when I right in a code
>
a + b
>
What really happened is that the language rewrote the code to
>
operator+(a,b)
>
and that because of it, it was possible to use the function notation
for all operators.
>
So is there anyway in which I can use the function notation instead of
the classic notation?

Like Ron suggested, use STL functors. In this case, something like this
might work:

#include <functional>

int main()
{
std::plus<intp;
int a=1,b=2;

int c = p(a,b);

return 0;
}