template; operator (int)

Question

regarding my Point struct already mentioned here:
template class: ctor against function -> new C++ standard
is there a chance to replace the function toint() with a cast-operator (int)?

namespace point { template < unsigned int dims, typename T > struct Point { T X[ dims ]; //umm??? template < typename U > Point< dims, U > operator U() const { Point< dims, U > ret; std::copy( X, X + dims, ret.X ); return ret; } //umm??? Point< dims, int > operator int() const { Point<dims, int> ret; std::copy( X, X + dims, ret.X ); return ret; } //OK Point<dims, int> toint() { Point<dims, int> ret; std::copy( X, X + dims, ret.X ); return ret; } }; //struct Point template < typename T > Point< 2, T > Create( T X0, T X1 ) { Point< 2, T > ret; ret.X[ 0 ] = X0; ret.X[ 1 ] = X1; return ret; } }; //namespace point int main(void) { using namespace point; Point< 2, double > p2d = point::Create( 12.3, 34.5 ); Point< 2, int > p2i = (int)p2d; //äähhm??? std::cout << p2d.str() << std::endl; char c; std::cin >> c; return 0; }

I think the problem is here that C++ cannot distinguish between different return types? many thanks in advance. regards
Oops

your syntax is wrong. do you want to cast to int? do you want to cast to different type point? — Anycorn
– Anycorn, Commented May 9, 2010 at 16:06

kennytm · Accepted Answer · 2010-05-09 16:04:55Z

The correct syntax is

 operator int() const { ...

There's no need to have that extra return type when you overload the cast operator.

And when you say (int)x, the compiler really expects to get an int, not a Point<dims, int>. Probably you want a constructor instead.

 template <typename U> Point(const Point<dims, U>& other) { ... }

Yes! thanks, this is probably the best solution. no other answer is needed.
Alternative correct syntax: operator Point<dims, int>() ;) But using a constructor for conversion is a better idea.

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