Template argument deduction

In order to instantiate a function template, every template argument must be known, but not every template argument has to be specified. When possible, the compiler will deduce the missing template arguments from the function arguments. This occurs when a function call is attempted, when an address of a function template is taken, and in some other contexts:

template<typename To, typename From> To convert(From f);
 
void g(double d) 
{
    int i = convert<int>(d); // calls convert<int,double>(double)
    char c = convert<char>(d); // calls convert<char,double>(double)
    int(*ptr)(float) = convert; // instantiates convert<int, float>(float)
}

This mechanism makes it possible to use template operators, since there is no syntax to specify template arguments for an operator other than by re-writing it as a function call expression.

#include <iostream>
int main() 
{
    std::cout << "Hello, world" << std::endl;
    // operator<< is looked up via ADL as std::operator<<,
    // then deduced to operator<<<char, std::char_traits<char>> both times
    // std::endl is deduced to &std::endl<char, std::char_traits<char>>
}

Template argument deduction takes place after the function template name lookup (which may involve argument-dependent lookup) and before template argument substitution (which may involve SFINAE) and overload resolution.

Deduction from a function call

Template argument deduction attempts to determine template arguments (types for type template parameters Ti, templates for template template parameters TTi, and values for non-type template parameters Ii), which can be substituted into each parameter P to produce the type deduced A, which is the same as the type of the argument A, after adjustments listed below.

If there are multiple parameters, each P/A pair is deduced separately and the deduced template arguments are then combined. If deduction fails or is ambiguous for any P/A pair or if different pairs yield different deduced template arguments, or if any template argument remains neither deduced nor explicitly specified, compilation fails.

If removing references and cv-qualifiers from P gives std::initializer_list<P'> and A is a braced-init-list, then deduction is performed for every element of the initializer list, taking P' as the parameter and the list element as the argument.

template<class T> void f(std::initializer_list<T>);
f({1,2,3});   // P=std::initializer_list<T>, A={1,2,3}
              // P'1=T, A'1=1: deduces T=int
              // P'2=T, A'2=2: deduces T=int
              // P'3=T, A'3=3: deduces T=int
              // deduction succeeds, T = int
f({1,"asdf"}); // P=std::initializer_list<T>, A={1,"asdf"}
               // P'1=T, A'1=1: deduces T=int
               // P'2=T, A'2="asdf", deduces T=const char*
               // deduction fails, T ambiguous

template<class T, int N> void h(T const(&)[N]);
h({1,2,3}); // T deduced to int, N deduced to 3
 
template<class T> void j(T const(&)[3]);
j({42});  // T deduced to int, array bound is not a parameter, not considered
 
struct Aggr { int i; int j; };
template<int N> void k(Aggr const(&)[N]);
k({1,2,3});       // error: deduction fails, no conversion from int to Aggr
k({{1},{2},{3}}); // OK, N deduced to 3
 
template<int M, int N> void m(int const(&)[M][N]);
m({{1,2},{3,4}});        // M and N both deduced to 2
 
template<class T, int N> void n(T const(&)[N], T);
n({{1},{2},{3}},Aggr()); // OK, T is Aggr, N is 3

If a parameter pack appears as the last P, then the type P is matched against the type A of each remaining argument of the call. Each match deduces the template arguments for the next position in the pack expansion.

template<class ... Types> void f(Types& ...);
void h(int x, float& y) {
   const int z = x;
   f(x, y, z); // P=Types&..., A1=x: deduces the first member of Types... to int
               // P=Types&..., A2=y: deduces the second member of Types... to float
               // P=Types&..., A3=z: deduces the third member of Types... to const int
               // calls f<int, float, const int>

If P is a function type, pointer to function type, or pointer to member function type and if A is a set of overloaded functions not containing function templates, template argument deduction is attempted with each overload. If only one succeeds, that successful deduction is used. If none or more than one succeeds, the template parameter is non-deduced context (see below).

template <class T> int f(T (*p)(T));
int g(int);
int g(char);
 
f(g); // P=T(*)(T), A=overload set
      // P=T(*)(T), A1=int(int): deduces T=int
      // P=T(*)(T), A2=int(char): fails to deduce T
      // only one overload works, deduction succeeds

Before deduction begins, the following adjustments to P and A are made:

template <class T> void f(T);
 
int a[3];
f(a); // P = T, A = int[3], adjusted to int*. Deduces T = int*
const int b = 13;
f(b); // P = T, A = const int, adjusted to int. Deduces T = int
void g(int);
f(g); // P = T, A = void(int), adjusted to void(*)(int). Deduces T = void(*)(int)

template <class T>
int f(T&&); // P is rvalue reference to cv-unqualified T ("forwarding reference")
 
template <class T>
int g(const T&&); // P is rvalue reference to cv-qualified T (not special)
 
int main()
{
    int i;
    int n1 = f(i); // argument is lvalue:     calls f<int&>(int&) (special case)
    int n2 = f(0); // argument is not lvalue: calls f<int>(int&&)
 
//  int n3 = g(i); // error: deduces to g<int>(const int&&), which
                   // cant bind an rvalue reference to an lvalue:
}

After these transformations, the deduction process as described below (in "Deduction from type"), and attempts to find such template arguments that would make deduced A (that is, P after adjustments listed above and the substitution of the deduced template parameters) identical to the transformed A, that is A after the adjustments listed above.

If the usual deduction from P and A fails, the following alternatives are additionally considered:

template<typename T> void f1(const T& t);
bool a = false;
f1(a); // P=const T&, adjusted to const T, A=bool, 
       // deduced T = bool, deduced A = const bool
       // deduced A is more cv-qualified than A

template<typename T> void f(const T*);
int* p;
f(p); // P=T, A=int*
      // deduces T=int, deduced A = const int*
      // qualification conversion applies (from int* to const int*)

template <class T> struct B { };
template <class T> struct D : public B<T> {};
 
template <class T> void f(B<T>&){}
 
void f() {
    D<int> d;
    f(d);  // P = B<T>&, adjusted P = B<T> (a simple-template-id)
           // A = D<int>
           // deduced T = int, deduced A = B<int>
           // A is derived from deduced A
}

Non-deduced contexts

In the following cases, the types, templates, and non-type values that are used to compose P do not participate in template argument deduction, but instead use the template arguments that were either deduced elsewhere or explicitly specified. If a template parameter is used only in non-deduced contexts and is not explicitly specified, template argument deduction fails.

// the identity template, often used to exclude specific arguments from deduction
template <typename T> struct identity { typedef T type; };
 
template <typename T>
void bad(std::vector<T> x, T value = 1);
 
template <typename T>
void good(std::vector<T> x, typename identity<T>::type value = 1);
 
std::vector<std::complex<double>> x;
bad(x, 1.2); // P1 = std::vector<T>, A1 = std::vector<std::complex<double>>
             // P1/A1 deduction determines T = std::complex<double>
             // P2 = T, A2 = double
             // P2/A2 deduction determines T = double -- Error
good(x, 1.2); // P1/A1 deduces T = std::complex<double>
              // P2 = identity<T>::type, T is to the left of ::, non-deduced

template<typename T> void f(decltype(*std::declval<T>()) arg);
int n; f<int*>(n); // P = decltype(*declval<T>()), A=int. T is not deducible

template<std::size_t N> void f(std::array<int, 2*N> a);
std::array<int, 10> a;
f(a); // P=std::array<int, 2*N>, "2*N" is non-deduced context, N cannot be deduced
      // (note: "f(std::array<int, N> a)" would be able to deduce N)

template<typename T, typename F>
void f(const std::vector<T> &arr, const F& comp = std::less<T>());
std::vector<std::string> arr(3);
f(arr); // P1 = const std::vector<T> &, A1=std::vector<std::string> lvalue,
        // P1/A1 deduces T = std::string
        // P2 = non-deduced context for F (template parameter) used in the
        // parameter type const F& of the function parameter comp,
        // that has a default argument that is being used in the call f(arr)

template<typename T> void out(const T& value) { std::cout << value; }
 
out("12345");   // P=const T&, A=const char[6] lvalue, T is deduced to be "char[6]"
out(std::endl); // P=const T&, A=function template, T is non-deducible, error.

template<class T> void g1(std::vector<T>);
g1({1,2,3}); // P=std::vector<T>, A={1,2,3}, this P is non-deduced context
             // Error: T is not explicitly specified or deduced from another P/A
 
template<class T> void g2(std::vector<T>, T x);
g2({1,2,3}, 10); // P1=std::vector<T,> A1={1,2,3}, this P is non-deduced context
                 // P2=T, A2=int, this P/A pair deduces T = int

template<class... Ts, class T> void f1(T n, Ts... args) {}
template<class... Ts, class T> void f2(Ts... args, T n) {}
 
f1(1,2,3,4); // P1=T, A1=1: T is deduced to be int
             // P2=Ts..., A2=2, A3=3, A4=4: Ts is deduced to be [int, int, int]
f2(1,2,3,4); // P1=Ts..., non-deducible

template<int...> struct T {};
 
template<int... Ts1, int N, int... Ts2>
void good(const T<N, Ts1...>& arg1, const T<N, Ts2...>&) {}
 
template<int... Ts1, int N, int... Ts2>
void bad(const T<Ts1..., N>& arg1, const T<Ts2..., N>&) {}
 
T<1,2> t1; T<1,-1,0> t2;
 
good(t1, t2); // P1=const T<N, Ts1...>&, A1=T<1,2>: deduces N=1, Ts1=[2]
              // P2=const T<N, Ts2...>&, A2=T<1,-1,0>: deduces N=1, Ts2=[-1, 0]
 
bad(t1, t2); // P1=const T<Ts1..., N>&, A1=T<1,2>:  <Ts1..., N> is non-deduced
             // P2=const T<Ts2..., N>&, A2=T<1,-1,0>: <Ts2..., N> is non-deduced

template<int i> void f1(int a[10][i]);
template<int i> void f2(int a[i][20]); // P = int[i][20], array type
template<int i> void f3(int (&a)[i][20]); // P =int(&)[i][20], reference to array
void g() {
  int v[10][20];
  f1(v);            // OK: i deduced to be 20
  f1<20>(v);        // OK
  f2(v);            // error: cannot deduce template-argument i
  f2<10>(v);        // OK
  f3(v);            // OK: i deduced to be 10
}

In any case, if any part of a type name is non-deduced context, the entire type name is non-deduced. However, compound types can include both deduced and non-deduced type names. For example, in A<T>::B<T2>, T is non-deduced because of rule #1 (nested name specifier), and T2 is non-deduced because it is part of the same type name, but in void(*f)(typename A<T>::B, A<T>), the T in A<T>::B is non-deduced (because of the same rule), while the T in A<T> is deduced.

Deduction from a type

Given a function parameter P that depends on one or more type template parameters Ti, template template parameters TTi, or non-type template parameters Ii, and the corresponding argument A, deduction takes place if P has one of the following forms:

• T
• cv-list T
• T*
• T&
• T&&
• T[integer-constant]
• class-template-name<T>
• type(T)
• T()
• T(T)
• T type::*
• type T::*
• T T::*
• T(type::*)()
• type(T::*)()
• type(type::*)(T)
• type(T::*)(T)
• T (type::*)(T)
• T (T::*)()
• T (T::*)(T)
• type[i]
• class-template-name<I>
• TT<T>
• TT<I>
• TT<>

where.

• (T) is a function parameter type list where at least one parameter type contains T
• () is a function parameter type list where no parameters contain T
• <T> is a template argument list where at least one argument contains T
• <I> is a template argument list where at least one argument contains I
• <> is a template argument list where no arguments contain T or I

If P has one of the forms that include a template parameter list <T> or <I>, then each element Pi of that template argument list is matched against the corresponding template argument Ai of its A. If the last Pi is a pack expansion, then its pattern is compared against each remaining argument in the template argument list of A. A trailing parameter pack that is not otherwise deduced, is deduced to an empty parameter pack.

If P has one of the forms that include a function parameter list (T), then each parameter Pi from that list is compared with the corresponding argument Ai from A's function parameter list. If the last Pi is a pack expansion, then its declarator is compared with each remaining Ai in the parameter type list of A.

Forms can be nested and processed recursively: X<int>(*)(char[6]) is an example of type(*)(T), where type is class-template-name<T> and T is type[i].

Template type argument cannot be deduced from the type of a non-type template argument.

template<typename T, T i> void f(double a[10][i]);
double v[10][20];
f(v); // P=double[10][i] A=double[10][20]:
      // i can be deduced to equal 20
      // but T cannot be deduced from the type of i

If a non-type template parameter is used in the parameter list, and the corresponding template argument is deduced, the type of the deduced template argument must match the type of the non-type template parameter exactly (except that cv-qualifiers are dropped, and except where the template argument is deduced from an array bound -- in that case any integral type is allowed, even bool though it would always become true).

template<int i> class A {};
template<short s> void f(A<s>); // the type of the non-type template param is short
void k1() {
  A<1> a; // the type of the non-type template param of a is int
  f(a);  // P=A<(short)s>, A=A<(int)1>, Error: deduced non-type template argument 
         // does not have the same type as its corresponding template argument
  f<1>(a); // OK, the template argument is not deduced, 
           // this calls f<(short)1>(A<(short)1>)
}

Template type parameter cannot be deduced from the type of a function default argument.

template<typename T> void f(T = 5, T = 7);
void g() {
  f(1);      // OK: calls f<int>(1,7)
  f();       // error: cannot deduce T
  f<int>();  // OK: calls f<int>(5,7)
}

Deduction of template template parameter can use the type used in the template specialization used in the function call.

template <template<typename> class X> struct A {}; // A is a template with a TT param
template <template<typename> class TT> void f(A<TT>) {} 
template<class T> struct B { }; 
A<B> ab;
f(ab); // P=A<TT>, A=A<B>. This deduces TT=B and calls f(A<B>)

Other contexts

Besides function call expressions and an operator expressions, template argument deduction is used in the following situations:

auto type deduction

Template argument deduction is used in declarations of variables, when deducing the meaning of the auto specifier from the variable's initializer.

The parameter P is obtained as follows: in T, the declared type of the variable that includes auto, every occurrence of auto is replaced with an imaginary type template parameter U or, if the initializer is a brace-init-list, with std::initializer_list<U>. The argument A is the initializer expression. After deduction of U from P and A following the rules described above, the deduced U is substituted into T to get the actual variable type.

const auto& x = 1+2; // P = const U&, A = 1+2
  // same rules as for calling f(1+2) where f is
  // template <class U> void f(const U& u)
  // U is deduced to be int, and the type of x is const int&
auto l = {13}; // P = std::initializer_list<U>, A = {13}
   // U is deduced to be int, the type of l is std::initializer_list<int>

auto x1 = {3}; // x1 is std::initializer_list<int>
auto x2{1, 2}; // error: not a single element
auto x3{3}; // x3 is int (before C++17 it was std::initializer_list<int>)

auto f() { return 42; } // P = auto, A = 42
 // U is deduced to be int, the return type of f is int

overload resolution

Template argument deduction is used during overload resolution, when generating specializations from a candidate template function. P and A are the same as in a regular function call.

std::string s;
std::getline(std::cin, s); // "std::getline" names 4 function templates, 
  // 2 of which are candidate functions (correct number of parameters)
  // 1st candidate template:
  // P1 = std::basic_istream<CharT,Traits>&, A1 = std::cin
  // P2 = std::basic_string<CharT,Traits,Allocator>&, A2 = s
  // deduction determines the type template parameters CharT, Traits, and Allocator
  // specialization std::getline<char, std::char_traits<char>, std::allocator<char>>
  // 2nd candidate template:
  // P1 = std::basic_istream<CharT,Traits>&&, A1 = std::cin
  // P2 = std::basic_string<CharT,Traits,Allocator>&, A2 = s
  // deduction determines the type template parameters CharT, Traits, and Allocator
  // specialization std::getline<char, std::char_traits<char>, std::allocator<char>>
  // overload resolution ranks reference binding from lvalue std::cin and picks
  // the first of the two candidate specializations

If deduction fails, or if deduction succeeds, but the specialization it produces would be invalid (for example, an overloaded operator whose parameters are neither class nor enumeration types), (since C++14) the specialization is not included in the overload set, similar to SFINAE.

address of an overload set

Template argument deduction is used when taking an address of a overload set, which includes function templates.

The function type of the function template is P. The target type is the type of A.

std::cout << std::endl; // std::endl names a function template
// type of endl P = std::basic_ostream<CharT,Traits>&(std::basic_ostream<CharT,Traits>&)
// operator<< parameter A = std::basic_ostream<char,std::char_traits<char>>&(*)(std::basic_ostream<char,std::char_traits<char>>&)
// (other overloads of operator<< are not viable) 
// deduction determines the type template parameters CharT and Traits

An additional rule is applied to the deduction in this case: when comparing function parameters Pi and Ai, if any Pi is an rvalue reference to cv-unqualified template parameter (a "forwarding reference") and the corresponding Ai is an lvalue reference, then Pi is adjusted to the template parameter type (T&& becomes T).

partial ordering

Template argument deduction is used during partial ordering of overloaded function templates.

conversion function template

Template argument deduction is used when selecting user-defined conversion function template arguments.

A is the type that is required as the result of the conversion. P is the return type of the conversion function template, except that.

If A is cv-qualified, the top-level cv qualifiers are ignored. If A is a reference type, the referred type is used by deduction.

If, the usual deduction from P and A (as described above) fails, the following alternatives are additionally considered:

struct A {
    template <class T> operator T***();
};
A a;
const int * const * const * p1 = a; // A = const int * const * const *
                                    // P = T***
 // regular function-call deduction for 
 // template<class T> void f(T*** p) as if called with the argument
 // of type const int * const * const * fails
 // additional deduction for conversion functions determines T = int 
 // (deduced A is int***, convertible to const int * const * const *)

See member template for other rules regarding conversion function templates.

explicit instantiation

Template argument deduction is used in explicit instantiations, explicit specializations, and those friend declarations where the declarator-id happens to refer to a specialization of a function template (for example, friend ostream& operator<< <> (...)), if not all template arguments are explicitly specified or defaulted, template argument deduction is used to determine which template's specialization is referred to.

P is the type of the function template that is being considered as a potential match, and A is the function type from the declaration. If there are no matches or more than one match (after partial ordering), the function declaration is ill-formed.

template<class X> void f(X a);  // first template f
template<class X> void f(X* a); // second template f
 
template<> void f<>(int *a) {} // explicit specialization of f
 // P1 = void(X),  A1=void(int*): deduces X=int*
 // P2 = void(X*), A2=void(int*): deduces X=int
 // foo<int*>(int*) and f<int>(int*) are then submitted to partial ordering
 // which selects f<int>(int*) as the more specialized

An additional rule is applied to the deduction in this case: when comparing function parameters Pi and Ai, if any Pi is an rvalue reference to cv-unqualified template parameter (a "forwarding reference") and the corresponding Ai is an lvalue reference, then Pi is adjusted to the template parameter type (T&& becomes T).

deallocation function template

Template argument deduction is used when determining if a deallocation function template specialization matches a given placement form of operator new.

P is the type of the function template that is being considered as a potential match, and A is the function type of the deallocation function that would be the match for the placement operator new under consideration. If there is no match or more than one match (after overload resolution), the placement deallocation function is not called (memory leak may occur).

struct X {
    X() { throw std::runtime_error(""); }
    static void* operator new(std::size_t sz, bool b) { return ::operator new(sz); }
    static void* operator new(std::size_t sz, double f) { return ::operator new(sz); }
    template<typename T> static void operator delete(void* ptr, T arg) {
        ::operator delete(ptr);
    }
};
int main() {
   try {
     X* p1 = new (true) X; // when X() throws, operator delete is looked up
                           // P1=void(void*, T), A1=void(void*, bool): deduces T=bool
                           // P2=void(void*, T), A2=void(void*, double): T=double
                           // overload resolution picks operator delete<bool>
   } catch(const std::exception&) { }
   try {
     X* p1 = new (13.2) X; // same lookup, picks operator delete<double>
   } catch(const std::exception&) { }
}

Alias templates

Alias templates are never deduced.

template<class T> struct Alloc {};
template<class T> using Vec = vector<T, Alloc<T>>;
Vec<int> v;
 
template<template<class,class> class TT> void g(TT<int, Alloc<int>>);
g(v); // ok: deduces TT = vector
 
template<template<class> class TT> void f(TT<int>);
f(v); // error: TT cannot be deduced as "Vec" because Vec is an alias template

Implicit conversions

Type deduction does not consider implicit conversions (other than type adjustments listed above): that's the job for overload resolution, which happens later.

template <class T> struct Z {
    typedef typename T::x xx;
};
 
template <class T> typename Z<T>::xx f(void*, T); // #1
template <class T> void f(int, T);                // #2
 
struct A {} a;
int main() {
  f(1, a);  // for #1, deduction determines T = struct A, but the remaining argument 1
            // cannot be implicitly converted to its parameter void*, so deduction fails
            // Instantiation of the return type is not requested.
            // for #2, deduction determines T = struct A, and the remaining argument 1
            // can be implicitly converted to its parameter int, deduction succeeds for #2
            // The function call compiles as a call to #2 (deduction failure is SFINAE)
}

Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.