From:

"Victor Bazarov" <v.Abazarov@comAcast.net>

Newsgroups:

comp.lang.c++

Date:

Wed, 28 Jun 2006 22:13:02 -0400

Message-ID:

<Eu2dneQpb-Ozqj7ZnZ2dnUVZ_qGdnZ2d@comcast.com>

I am having trouble with overloading the += operator when template

parameters are used. I have a class holding an array (called

"derived" in the following example) which derives from a base class

("base"). I want to be able to add:

1) any derived array holding class to any other derived array holding

class

2) any derived array holding class to a literal value (e.g int,

double, etc) for which addition is defined for the type in the array.

I would like both + and += operators and for appropriate casting to be

handled. A minimal example of what I am trying to do is below (note

that the derived class holds only one value for simplicity).

The two + operators work fine with GCC (versions 3.4 and 4.1). But

the two += don't work when defined together (they work fine when

defined one at a time). When performing derived<L> += derived<R> the

operator+=(base<L,D>, R) is selected rather than

operator+=(base<L,LD>, base<R,RD>). Borland C++ Builder has this

problem and a similar one with the the binary addtion operator. I am

really confused about GCC because the operator prototypes are

practically the same, so it would seem that the the problem += and +

operators are either both correct or wrong.

Using operator+=(base<L,D>, L) works, but without automatic casting I

am after. Any suggestions on how to get around this problem, at least

in GCC would be great!

Example program (88 lines):

#include <iostream>

// curiously recurring base class

template <typename T, typename D>

class base

{

public:

T& operator[](std::size_t i) { return static_cast<D&>(*this)[i]; }

const T& operator[](std::size_t i) const { return static_cast<const

D&>(*this)[i]; }

};

// curiously recurring derived class

template <typename T>

class derived : public base<T, derived<T> >

{

public:

derived(T t) : t_(t) {}

T& operator[](std::size_t i) {return t_; }

const T& operator[](std::size_t i) const {return t_; }

private:

T t_;

};

// type promotion trait

template <typename L, typename R>

class promote

{

};

template<>

class promote<int, double>

{

public:

typedef double type;

};

// binary addition operators with casting

template <typename L, typename D, typename R>

derived<typename promote<L,R>::type>

operator+ (const base<L, D>& l, const R& r)

{

std::cout << "base + literal" << std::endl;

typedef typename promote<L,R>::type P;

return derived<P>(static_cast<P>(l[0]) + static_cast<P>(r));

}

template <typename L, typename DL, typename R, typename DR>

derived<typename promote<L,R>::type>

operator+ (const base<L, DL>& l, const base<R, DR>& r)

{

std::cout << "base + base" << std::endl;

typedef typename promote<L,R>::type P;

return derived<P>(static_cast<P>(l[0]) + static_cast<P>(r[0]));

}

// self-assigned addition

template <typename L, typename D, typename R>

void

operator+= (base<L, D>& l, const R& r)

{

std::cout << "base += literal" << std::endl;

l[0] += static_cast<L>(r);

parameters are used. I have a class holding an array (called

"derived" in the following example) which derives from a base class

("base"). I want to be able to add:

1) any derived array holding class to any other derived array holding

class

2) any derived array holding class to a literal value (e.g int,

double, etc) for which addition is defined for the type in the array.

I would like both + and += operators and for appropriate casting to be

handled. A minimal example of what I am trying to do is below (note

that the derived class holds only one value for simplicity).

The two + operators work fine with GCC (versions 3.4 and 4.1). But

the two += don't work when defined together (they work fine when

defined one at a time). When performing derived<L> += derived<R> the

operator+=(base<L,D>, R) is selected rather than

operator+=(base<L,LD>, base<R,RD>). Borland C++ Builder has this

problem and a similar one with the the binary addtion operator. I am

really confused about GCC because the operator prototypes are

practically the same, so it would seem that the the problem += and +

operators are either both correct or wrong.

Using operator+=(base<L,D>, L) works, but without automatic casting I

am after. Any suggestions on how to get around this problem, at least

in GCC would be great!

Example program (88 lines):

#include <iostream>

// curiously recurring base class

template <typename T, typename D>

class base

{

public:

T& operator[](std::size_t i) { return static_cast<D&>(*this)[i]; }

const T& operator[](std::size_t i) const { return static_cast<const

D&>(*this)[i]; }

};

// curiously recurring derived class

template <typename T>

class derived : public base<T, derived<T> >

{

public:

derived(T t) : t_(t) {}

T& operator[](std::size_t i) {return t_; }

const T& operator[](std::size_t i) const {return t_; }

private:

T t_;

};

// type promotion trait

template <typename L, typename R>

class promote

{

};

template<>

class promote<int, double>

{

public:

typedef double type;

};

// binary addition operators with casting

template <typename L, typename D, typename R>

derived<typename promote<L,R>::type>

operator+ (const base<L, D>& l, const R& r)

{

std::cout << "base + literal" << std::endl;

typedef typename promote<L,R>::type P;

return derived<P>(static_cast<P>(l[0]) + static_cast<P>(r));

}

template <typename L, typename DL, typename R, typename DR>

derived<typename promote<L,R>::type>

operator+ (const base<L, DL>& l, const base<R, DR>& r)

{

std::cout << "base + base" << std::endl;

typedef typename promote<L,R>::type P;

return derived<P>(static_cast<P>(l[0]) + static_cast<P>(r[0]));

}

// self-assigned addition

template <typename L, typename D, typename R>

void

operator+= (base<L, D>& l, const R& r)

{

std::cout << "base += literal" << std::endl;

l[0] += static_cast<L>(r);

This doesn't compile for me when instantiated from

'd2 += d3'

'R' is 'derived<double>', 'L' is 'double'. There is no conversion

from 'derived<double>' to 'double'.

The existence of the other operator+= does not matter here, I am

guessing. The reason is that with 'd3' as the second argument

the compiler cannot deduce 'R' and 'DR' from 'derived<double>'.

return;

}

template <typename L, typename DL, typename R, typename DR>

void

operator+= (base<L, DL>& l, const base<R, DR>& r)

{

std::cout << "base += base" << std::endl;

l[0] += static_cast<L>(r[0]);

return;

}

int main()

{

derived<int> d1(2);

derived<double> d2 = d1 + 3.; // base + literal

derived<double> d3 = d1 + d2; // base + base

d3 += 3.; // base += literal

d2 += d3; // base += base

}

}

template <typename L, typename DL, typename R, typename DR>

void

operator+= (base<L, DL>& l, const base<R, DR>& r)

{

std::cout << "base += base" << std::endl;

l[0] += static_cast<L>(r[0]);

return;

}

int main()

{

derived<int> d1(2);

derived<double> d2 = d1 + 3.; // base + literal

derived<double> d3 = d1 + d2; // base + base

d3 += 3.; // base += literal

d2 += d3; // base += base

}

V

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