Why aren't operator conversions implicitly called for templated functions? (C++)
- by John Gordon
I have the following code:
template <class T>
struct pointer
{
operator pointer<const T>() const;
};
void f(pointer<const float>);
template <typename U>
void tf(pointer<const float>);
void g()
{
pointer<float> ptr;
f(ptr);
tf(ptr);
}
When I compile the code with gcc 4.3.3 I get a message (aaa.cc:17: error: no matching function for call to ‘tf(pointer<float>&)’) indicating that the compiler called 'operator pointer<const T>' for the non-templated function f(), but didn't for the templated function tf(). Why and is there any workaround short of overloading tf() with a const and non-const version?
Thanks in advance for any help.
