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J W
Guest
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 Posted: Fri Oct 28, 2005 4:04 pm Post subject: Can operator-> be a template function? |
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Is it legal for operator-> to be a template function? And if so, how do
you
specify the template parameter?
Example:
struct foo
{
template<int i>
foo* operator->() { return this; }
void myfunc() { }
};
foo f;
f-><1>myfunc();
This is a contrived example, in reality i want to have operator-> be a
template function and have specializations on the int param so i can
dictate
different behavior.
GCC 3.4.4 parses the templated operator-> but I cannot figure out how to
call it, as doing f-><1>myfunc(); does not work.
Jeff
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acehreli@yahoo.com
Guest
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| Posted: Sat Oct 29, 2005 7:39 pm Post subject: Re: Can operator-> be a template function? |
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J W wrote:
| Quote: | Is it legal for operator-> to be a template function? And if so, how do
you
specify the template parameter?
Example:
struct foo
{
template<int i
foo* operator->() { return this; }
void myfunc() { }
};
foo f;
f-><1>myfunc();
This is a contrived example, in reality i want to have operator-> be a
template function and have specializations on the int param so i can
dictate
different behavior.
|
I seriously think that there are better ways of doing what you actually
need, but here is what I was able to run:
#include <iostream>
using std::cout;
struct foo
{
template<int i>
foo * operator->()
{
cout << "defaultn";
return this;
}
void my_func()
{}
int i_;
char c_;
};
template <>
foo * foo::operator-> <42> ()
{
cout << "forty twon";
return this;
}
template <>
foo * foo::operator-> <7> ()
{
cout << "sevenn";
return this;
}
int main()
{
foo f;
foo * p = &f;
p->my_func();
p->operator-><42>()->my_func();
p->operator-><7>()->my_func();
p->operator-><0>()->my_func();
}
Ali
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jeff_d_harper@hotmail.com
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| Posted: Sat Oct 29, 2005 7:43 pm Post subject: Re: Can operator-> be a template function? |
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| Quote: | GCC 3.4.4 parses the templated operator-> but I cannot figure out how to
|
call it f.operator-><3>()->myfunc();
:-(
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Allan W
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| Posted: Sat Oct 29, 2005 7:45 pm Post subject: Re: Can operator-> be a template function? |
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J W wrote:
| Quote: | Is it legal for operator-> to be a template function?
|
Yes, I think so.
| Quote: | And if so, how do you specify the template parameter?
struct foo {
template<int i
foo* operator->() { return this; }
void myfunc() { }
};
foo f;
f-><1>myfunc(); // Doesn't work
|
This seems to work, at least on Visual Studio .Net 2003:
f.operator-> <1> () -> myfunc();
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Bronek Kozicki
Guest
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| Posted: Sun Oct 30, 2005 3:01 pm Post subject: Re: Can operator-> be a template function? |
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J W wrote:
| Quote: | This is a contrived example, in reality i want to have operator-> be a
template function and have specializations on the int param so i can
dictate different behavior.
|
As you already know, the syntax to call specialized operator is quite
ugly. If you do not insists on using numbers, you might consider
following syntax "f(foo::gaa)->myfunc();". Full example follows.
B.
#include<cstdio>
// foo.hpp
class foo
{
struct gaa_t {};
struct bar_t {};
public:
static const gaa_t * const gaa;
static const bar_t * const bar;
foo* operator()(const gaa_t *);
foo* operator()(const bar_t *);
void myfunc() { }
};
// foo.cpp
// include "foo.hpp"
foo* foo::operator()(const gaa_t *)
{std::puts("gaa"); return this;}
foo* foo::operator()(const bar_t *)
{std::puts("bar"); return this;}
const foo::gaa_t * const foo::gaa = 0;
const foo::bar_t * const foo::bar = 0;
// main.cpp
// include "foo.hpp"
int main()
{
foo f;
f(foo::gaa)->myfunc();
f(foo::bar)->myfunc();
}
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