Template argument deduction (C++ only)

When you call a template function, you may omit any template argument that the compiler can determine or deduce by the usage and context of that template function call.

The compiler tries to deduce a template argument by comparing the type of the corresponding template parameter with the type of the argument used in the function call. The two types that the compiler compares (the template parameter and the argument used in the function call) must be of a certain structure in order for template argument deduction to work. The following lists these type structures:

T
const T
volatile T
T&
T*
T[10]
A<T>
C(*)(T)
T(*)()
T(*)(U)
T C::*
C T::*
T U::*
T (C::*)()
C (T::*)()
D (C::*)(T)
C (T::*)(U)
T (C::*)(U)
T (U::*)()
T (U::*)(V)
E[10][i]
B<i>
TT<T>
TT<i>
TT<C>

The following example demonstrates the use of each of these type structures. The example declares a template function using each of the above structures as an argument. These functions are then called (without template arguments) in order of declaration. The example outputs the same list of type structures:

#include <iostream>
using namespace std;

template<class T> class A { };
template<int i> class B { };

class C {
   public:
      int x;
};

class D {
   public:
      C y;
      int z;
};

template<class T> void f (T)          { cout << "T" << endl; };
template<class T> void f1(const T)    { cout << "const T" << endl; };
template<class T> void f2(volatile T) { cout << "volatile T" << endl;  };
template<class T> void g (T*)         { cout << "T*" << endl; };
template<class T> void g (T&)         { cout << "T&" << endl; };
template<class T> void g1(T[10])      { cout << "T[10]" << endl;};
template<class T> void h1(A<T>)       { cout << "A<T>" << endl; };

void test_1() {
   A<char> a;
   C c;

   f(c);   f1(c);   f2(c);
   g(c);   g(&c);   g1(&c);
   h1(a);
}

template<class T>          void j(C(*)(T)) { cout << "C(*) (T)" << endl; };
template<class T>          void j(T(*)())  { cout << "T(*) ()" << endl; }
template<class T, class U> void j(T(*)(U)) { cout << "T(*) (U)" << endl; };

void test_2() {
   C (*c_pfunct1)(int);
   C (*c_pfunct2)(void);
   int (*c_pfunct3)(int);
   j(c_pfunct1);
   j(c_pfunct2);
   j(c_pfunct3);
}

template<class T>          void k(T C::*) { cout << "T C::*" << endl; };
template<class T>          void k(C T::*) { cout << "C T::*" << endl; };
template<class T, class U> void k(T U::*) { cout << "T U::*" << endl; };

void test_3() {
   k(&C::x);
   k(&D::y);
   k(&D::z);
}

template<class T>     void m(T (C::*)() )
   { cout << "T (C::*)()" << endl; };
template<class T>     void m(C (T::*)() )
   { cout << "C (T::*)()" << endl; };
template<class T>     void m(D (C::*)(T))
   { cout << "D (C::*)(T)" << endl; };
template<class T, class U>  void m(C (T::*)(U))
   { cout << "C (T::*)(U)" << endl; };
template<class T, class U>  void m(T (C::*)(U))
   { cout << "T (C::*)(U)" << endl; };
template<class T, class U>  void m(T (U::*)() )
   { cout << "T (U::*)()" << endl; };
template<class T, class U, class V> void m(T (U::*)(V))
   { cout << "T (U::*)(V)" << endl; };

void test_4() {
   int (C::*f_membp1)(void);
   C (D::*f_membp2)(void);
   D (C::*f_membp3)(int);
   m(f_membp1);
   m(f_membp2);
   m(f_membp3);

   C (D::*f_membp4)(int);
   int (C::*f_membp5)(int);
   int (D::*f_membp6)(void);
   m(f_membp4);
   m(f_membp5);
   m(f_membp6);

   int (D::*f_membp7)(int);
   m(f_membp7);
}

template<int i> void n(C[10][i]) { cout << "E[10][i]" << endl; };
template<int i> void n(B<i>)     { cout << "B<i>" << endl; };

void test_5() {
   C array[10][20];
   n(array);
   B<20> b;
   n(b);
}

template<template<class> class TT, class T> void p1(TT<T>)
   { cout << "TT<T>" << endl; };
template<template<int> class TT, int i>     void p2(TT<i>)
   { cout << "TT<i>" << endl; };
template<template<class> class TT>          void p3(TT<C>)
   { cout << "TT<C>" << endl; };

void test_6() {
   A<char> a;
   B<20> b;
   A<C> c;
   p1(a);
   p2(b);
   p3(c);
}

int main() { test_1(); test_2(); test_3(); test_4(); test_5(); test_6(); }

Deducing type template arguments

The compiler can deduce template arguments from a type composed of several of the listed type structures. The following example demonstrates template argument deduction for a type composed of several type structures:

template<class T> class Y { };

template<class T, int i> class X {
   public:
      Y<T> f(char[20][i]) { return x; };
      Y<T> x;
};

template<template<class> class T, class U, class V, class W, int i>
   void g( T<U> (V::*)(W[20][i]) ) { };

int main()
{
   Y<int> (X<int, 20>::*p)(char[20][20]) = &X<int, 20>::f;
   g(p);
}

The type Y<int> (X<int, 20>::*p)(char[20][20])T<U> (V::*)(W[20][i]) is based on the type structure T (U::*)(V):

If you qualify a type with the class to which that type belongs, and that class (a nested name specifier) depends on a template parameter, the compiler will not deduce a template argument for that parameter. If a type contains a template argument that cannot be deduced for this reason, all template arguments in that type will not be deduced. The following example demonstrates this:

template<class T, class U, class V>
  void h(typename Y<T>::template Z<U>, Y<T>, Y<V>) { };

int main() {
  Y<int>::Z<char> a;
  Y<int> b;
  Y<float> c;

  h<int, char, float>(a, b, c);
  h<int, char>(a, b, c);
  // h<int>(a, b, c);
}

The compiler will not deduce the template arguments T and U in typename Y<T>::template Z<U> (but it will deduce the T in Y<T>). The compiler would not allow the template function call h<int>(a, b, c) because U is not deduced by the compiler.

The compiler can deduce a function template argument from a pointer to function or pointer to member function argument given several overloaded function names. However, none of the overloaded functions may be function templates, nor can more than one overloaded function match the required type. The following example demonstrates this:

template<class T> void f(void(*) (T,int)) { };

template<class T> void g1(T, int) { };

void g2(int, int) { };
void g2(char, int) { };

void g3(int, int, int) { };
void g3(float, int) { };

int main() {
//   f(&g1);
//   f(&g2);
   f(&g3);
}

The compiler would not allow the call f(&g1) because g1() is a function template. The compiler would not allow the call f(&g2) because both functions named g2() match the type required by f().

The compiler cannot deduce a template argument from the type of a default argument. The following example demonstrates this:

template<class T> void f(T = 2, T = 3) { };

int main() {
   f(6);
//   f();
   f<int>();
}

The compiler allows the call f(6) because the compiler deduces the template argument (int) by the value of the function call's argument. The compiler would not allow the call f() because the compiler cannot deduce template argument from the default arguments of f().

The compiler cannot deduce a template type argument from the type of a non-type template argument. For example, the compiler will not allow the following:

template<class T, T i> void f(int[20][i]) { };

int main() {
   int a[20][30];
   f(a);
}

The compiler cannot deduce the type of template parameter T.

Deducing non-type template arguments

The compiler cannot deduce the value of a major array bound unless the bound refers to a reference or pointer type. Major array bounds are not part of function parameter types. The following code demonstrates this:

template<int i> void f(int a[10][i]) { };
template<int i> void g(int a[i]) { };
template<int i> void h(int (&a)[i]) { };

int main () {
   int b[10][20];
   int c[10];
   f(b);
   // g(c);
   h(c);
}

The compiler would not allow the call g(c); the compiler cannot deduce template argument i.

The compiler cannot deduce the value of a non-type template argument used in an expression in the template function's parameter list. The following example demonstrates this:

template<int i> class X { };

template<int i> void f(X<i - 1>) { };

int main () {
  X<0> a;
  f<1>(a);
  // f(a);
}

In order to call function f() with object a, the function must accept an argument of type X<0>. However, the compiler cannot deduce that the template argument i must be equal to 1 in order for the function template argument type X<i - 1> to be equivalent to X<0>. Therefore the compiler would not allow the function call f(a).

If you want the compiler to deduce a non-type template argument, the type of the parameter must match exactly the type of value used in the function call. For example, the compiler will not allow the following:

template<int i> class A { };
template<short d> void f(A<d>) { };

int main() {
   A<1> a;
   f(a);
}

The compiler will not convert int to short when the example calls f().

However, deduced array bounds may be of any integral type.