template parameter and conversion operator

[jacek]

You could try:

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

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

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

[bitChanger]

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.

[jacek]

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:

#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;
}

#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;
}

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Output:

$ ./a.out
double

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

[bitChanger]

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?

#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;
}

#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;
}

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[bood]

type = val.operator ValType();

type = val.operator ValType();

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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...

type = static_cast<ValType>(val);

type = static_cast<ValType>(val);

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

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...

[bitChanger]

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.

[jacek]

Then maybe something like this:

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 >();
}

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 >();
}

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