In function call, why doesn't nullptr match a pointer to a template object?

Question

Here is an example of a code that works perfectly:

---

#include
#include

template< class D, template< class D, class A >         


        

Answer 1

This is just how template deduction work: no conversion take place.

The problem is not endemic to nullptr either, consider the extremely simple case:

#include 

struct Thing {
    operator int() const { return 0; }
} thingy;

template 
void print(T l, T r) { std::cout << l << " " << r << "\n"; }

int main() {
    int i = 0;
    print(i, thingy);
    return 0;
}

which yields:

prog.cpp: In function ‘int main()’:
prog.cpp:12:17: error: no matching function for call to ‘print(int&, Thing&)’
  print(i, thingy);
                 ^
prog.cpp:12:17: note: candidate is:
prog.cpp:8:6: note: template void print(T, T)
 void print(T l, T r) { std::cout << l << " " << r << "\n"; }
      ^
prog.cpp:8:6: note:   template argument deduction/substitution failed:
prog.cpp:12:17: note:   deduced conflicting types for parameter ‘T’ (‘int’ and ‘Thing’)
  print(i, thingy);
                 ^

Thus, the conversion of nullptr to int* does not occur prior to template argument deduction either. As mentioned, you have two ways of solving the issue:

• specifying the template parameters (thus no deduction occur)
• converting the argument yourself (deduction occurs, but after your explicit conversion)