Re: overload "operator->"

=?ISO-8859-1?Q?Erik_Wikstr=F6m?= <>
Sun, 06 May 2007 11:59:55 GMT
On 2007-05-06 13:22, Jess wrote:

On May 4, 7:32 pm, James Kanze <> wrote:

You can't change the syntax of C++ with operator overloading.
You can only extend it to support additional types.

By "extending", I guess I can't change a postfix to a prefix, or its


Can I change a member function to a non-member and vice

No, but for most(?) operators there's both a member and a non-member

Can I change a unary operator to a binary or a binary to a


Also, when ">>" and "<<" are used by iostream, are then member
operators of iostream or non-member functions?

Non-member, and that is how you should make the if you ever want to add
an operator<< to your class so that you can easily print it.

As far as I know the only operator that you have some freedom with is
the ()-operator, with which you can decide how many argument it takes,
all the others are quite fixed, in arity, precedence, postfix/prefix, etc.

b. why is b->f() the shorthand of
(b.operator->()) -> f()

Because that's what the standard says.

I see, it's the standard, though it doesn't look very obvious why
standard says it. :) When "->" is used with a pointer "p->b()" is
equivalent to (*p).b(). which doesn't show how we can use "->" with
user-defined class types.

The idea is that you should be able to make a class that acts just like
a pointer, and that requires that it works like that, take a look at
this code and compare my Pointer-class with a real pointer:

#include <string>
#include <iostream>

template<class T>
class Pointer
    T* ptr_;
    Pointer(T* p) : ptr_(p) {}
    T* operator->() { return ptr_; }
    T& operator*() { return *ptr_; }

int main()
    std::cout << "Using Pointer:\n";
    Pointer<std::string> P(new std::string());
    *P = "Hello";
    P->append(" World\n");
    std::cout << *P;

    std::cout << "Using a std::string pointer:\n";
    std::string* p = new std::string();
    *p = "Hello";
    p->append(" World\n");
    std::cout << *p;

Erik Wikstr?m

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