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Erik Wikström
Guest
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 Posted: Fri Dec 15, 2006 10:11 am Post subject: Pointer to operator for builtin types |
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Is there some way to get a function-pointer to the operators of the
builtin types? In the following example I have a generic function which
applies an operator on two values, however it does not work on builtin
types, is there a way to make this work?
struct Test
{
int i;
Test(int i_) : i(i_) { }
Test operator+(Test& t) { return Test(i + t.i); }
Test operator-(Test& t) { return Test(i - t.i); }
operator int() { return i; }
};
template<class T>
T operate(T& t1, T& t2, T (T::*func)(T&))
{
return (t1.*func)(t2);
}
int main() {
Test test1(1);
Test test2(2);
Test test3 = operate(test1, test2, &Test::operator+);
Test test4 = operate(test1, test2, &Test::operator-);
/*
int i1 = 1;
int i2 = 2;
int i3 = operate(i1, i2, ???);
int i4 = operate(i1, i2, ???)
*/
}
--
Erik Wikström |
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Ondra Holub
Guest
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| Posted: Fri Dec 15, 2006 10:11 am Post subject: Re: Pointer to operator for builtin types |
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Erik Wikström napsal:
| Quote: | Is there some way to get a function-pointer to the operators of the
builtin types? In the following example I have a generic function which
applies an operator on two values, however it does not work on builtin
types, is there a way to make this work?
struct Test
{
int i;
Test(int i_) : i(i_) { }
Test operator+(Test& t) { return Test(i + t.i); }
Test operator-(Test& t) { return Test(i - t.i); }
operator int() { return i; }
};
template<class T
T operate(T& t1, T& t2, T (T::*func)(T&))
{
return (t1.*func)(t2);
}
int main() {
Test test1(1);
Test test2(2);
Test test3 = operate(test1, test2, &Test::operator+);
Test test4 = operate(test1, test2, &Test::operator-);
/*
int i1 = 1;
int i2 = 2;
int i3 = operate(i1, i2, ???);
int i4 = operate(i1, i2, ???)
*/
}
--
Erik Wikström
|
AFAIK you cannot get pointer to operator+ for built-in types. But you
can wrap such type in some class:
#include <iostream>
struct Test
{
int i;
Test(int i_) : i(i_)
{ }
Test operator+(const Test& t)
{
return Test(i + t.i);
}
Test operator-(const Test& t)
{
return Test(i - t.i);
}
operator int()
{
return i;
}
};
template<typename T>
class WrapType
{
public:
WrapType(const T& data)
: data_(data)
{
}
T operator+(const T& value)
{
return data_ + value;
}
T operator-(const T& value)
{
return data_ - value;
}
private:
T data_;
};
template<typename T, typename U>
T operate(const T& t1, const T& t2, T (U::*func)(const T&))
{
return (U(t1).*func)(t2);
}
int main()
{
Test test1(1);
Test test2(2);
Test test3 = operate(test1, test2, &Test::operator+);
Test test4 = operate(test1, test2, &Test::operator-);
int i1 = 1;
int i2 = 2;
int i3 = operate(i1, i2, &WrapType<int>::operator+);
int i4 = operate(i1, i2, &WrapType<int>::operator-);
std::cout << i3 << " " << i4 << "\n";
}
If you need to call operate from some other template, there is no
problem to use WrapType always:
Test test3 = operate(test1, test2, &Wrap<Test>::operator+);
Of course, you would need to define also other operators in a WrapType
template. |
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John Carson
Guest
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| Posted: Fri Dec 15, 2006 10:11 am Post subject: Re: Pointer to operator for builtin types |
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"Erik Wikström" <eriwik (AT) student (DOT) chalmers.se> wrote in message
news:1166174015.343152.236160 (AT) 80g2000cwy (DOT) googlegroups.com
| Quote: | Is there some way to get a function-pointer to the operators of the
builtin types?
|
No.
| Quote: | In the following example I have a generic function
which applies an operator on two values, however it does not work on
builtin types, is there a way to make this work?
struct Test
{
int i;
Test(int i_) : i(i_) { }
Test operator+(Test& t) { return Test(i + t.i); }
Test operator-(Test& t) { return Test(i - t.i); }
operator int() { return i; }
};
template<class T
T operate(T& t1, T& t2, T (T::*func)(T&))
{
return (t1.*func)(t2);
}
int main() {
Test test1(1);
Test test2(2);
Test test3 = operate(test1, test2, &Test::operator+);
Test test4 = operate(test1, test2, &Test::operator-);
/*
int i1 = 1;
int i2 = 2;
int i3 = operate(i1, i2, ???);
int i4 = operate(i1, i2, ???)
*/
}
|
Why do you want to call operator+? Why not just use +. This works for both
built-in and class objects. The trick is to make class objects work like
built-in ones, not the reverse.
--
John Carson |
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