template parameter and conversion operator

The following code will compile with MSVC 7.1 and with g++ 2.95.3
However, I need it to compile with g++ 3.4.4 which it does not.

If i'm reading my books correctly the template parameter has infinite possible types while operator conversions have a finite number of types, therefore this breaks the c++ standard. :o

Can someone help me find a workaround and/or a g++ flag to allow this to work.

Thanks.

g++ 3.x and up give the following error:

main.cpp: In function `void setType(const MyType&)':
main.cpp:21: error: 'const class MyType' has no member named 'operator ValType'

Code:

---

#include <string>
using namespace std;
 
class MyType
{
public:
  MyType() {}
  ~MyType() {}
 
  operator char() const { return 'a'; }
  operator int() const { return 1; }
  operator double() const { return 2.0; }
  operator float() const { return 3.0f; }
};
 
template<typename ValType>
void setType(const MyType& val)
{
  ValType type;
  type = val.operator ValType();
}
 
int main()
{
  MyType a;
  setType<double>(a);
  return 0;
}

---

You could try:

Code:

---

template< typename ValType >
void setType( const MyType& val )
{
  ValType type;
  type = static_cast< ValType >( val );
}

---

PS. I use g++ 3.3.6 and both versions compile fine.

I've tried to compile it with 3.4.4 and 4.1.0 and got the same error.

In my accual application I can't just do a static cast on val because the conversion operator actually converts a private void* of data into the appropriate type not the class itself.

Quote:

---

Originally Posted by bitChanger

In my accual application I can't just do a static cast on val because the conversion operator actually converts a private void* of data into the appropriate type not the class itself.

---

It does exactly the same thing:

Code:

---

#include <iostream>
#include <string>
using namespace std;
 
class MyType
{
   public:
      MyType() {}
      ~MyType() {}
 
      operator char() const { cerr << "char" << endl; return 'a'; }
      operator int() const { cerr << "int" << endl; return 1; }
      operator double() const { cerr << "double" << endl; return 2.0; }
      operator float() const { cerr << "float" << endl; return 3.0f; }
};
 
template<typename ValType>
void setType(const MyType& val)
{
   ValType type;
   type = static_cast< ValType> ( val );
}
 
int main()
{
   MyType a;
   setType<double>(a);
   return 0;
}

---

Output:

Quote:

---

$ ./a.out
double

---

You could implement it in a similar way as QVariant::value() was implemented.

Now I know why I was trying to explicitly call the conversion operator. In my application the type is complex. This gets ambiguous with the complex copy constructor.

Any thoughts on this one? ;)

Code:

---

#include <string>
#include <complex>
using namespace std;
 
class MyType
{
public:
  MyType() {}
  ~MyType() {}
 
  operator char() const { return 'a'; }
  operator int() const { return 1; }
  operator double() const { return 2.0; }
  operator float() const { return 3.0f; }
  operator complex<double>() const { return complex<double>(1.0,1.0); }
};
 
template<typename ValType>
void setType(const MyType& val)
{
  ValType type;
  type = static_cast<ValType>(val);
}
 
int main()
{
  MyType a;
  setType<complex<double> >(a);
  return 0;
}

---

Code:

---

type = val.operator ValType();

---

Codes above does not work coz `ValType` here is a qulified name, so it's looked up only in the scope of val, i.e. MyType, so the `typename ValType` declared in the template function won't be looked up. But I haven't seen any solutions by now...:(

Code:

---

type = static_cast<ValType>(val);

---

However, it's weired that gcc3.4.2 reports an ambiguous. Gcc says there are two candidates:

Quote:

---

std::complex<double>::complex(const std::complex<double>&)
and
std::complex<double>::complex(double, double)

---

But the second one has two arguments, how can it beambiguous? I cannot understand why...

If you look at the 2 candidates that it provides you'll notice that they are constructors for the std::complex class. The compiler is confused and don't even see the conversion operator in this case.

Then maybe something like this:

Code:

---

class MyType
{
   public:
        MyType() {}
       ~MyType() {}
 
       template< typename T > T convertTo() const;
};
 
template<>
char MyType::convertTo< char >() const { return 'a'; }
 
template<>
int MyType::convertTo< int >() const { return 1; }
 
...
 
template<>
complex< double > MyType::convertTo< complex< double > >() const
{ 
   return complex< double >( 1.0, 1.0 );
}
 
template< typename ValType >
void setType( const MyType& val )
{
   ValType type;
   type = val.template convertTo< ValType >();
}

---