template; operator (int)
Posted
by Oops
on Stack Overflow
See other posts from Stack Overflow
or by Oops
Published on 2010-05-09T16:02:16Z
Indexed on
2010/05/09
16:08 UTC
Read the original article
Hit count: 166
Hi,
regarding my Point struct already mentioned here:
http://stackoverflow.com/questions/2794369/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
© Stack Overflow or respective owner