17 Templates [temp]

17.8 Function template specializations [temp.fct.spec]

17.8.1 Explicit template argument specification [temp.arg.explicit]

Template arguments can be specified when referring to a function template specialization by qualifying the function template name with the list of template-arguments in the same way as template-arguments are specified in uses of a class template specialization. [Example:

template<class T> void sort(Array<T>& v);
void f(Array<dcomplex>& cv, Array<int>& ci) {
  sort<dcomplex>(cv);                   // sort(Array<dcomplex>&)
  sort<int>(ci);                        // sort(Array<int>&)


template<class U, class V> U convert(V v);

void g(double d) {
  int i = convert<int,double>(d);       // int convert(double)
  char c = convert<char,double>(d);     // char convert(double)

end example]

A template argument list may be specified when referring to a specialization of a function template

Trailing template arguments that can be deduced or obtained from default template-arguments may be omitted from the list of explicit template-arguments. A trailing template parameter pack not otherwise deduced will be deduced to an empty sequence of template arguments. If all of the template arguments can be deduced, they may all be omitted; in this case, the empty template argument list <> itself may also be omitted. In contexts where deduction is done and fails, or in contexts where deduction is not done, if a template argument list is specified and it, along with any default template arguments, identifies a single function template specialization, then the template-id is an lvalue for the function template specialization. [Example:

template<class X, class Y> X f(Y);
template<class X, class Y, class ... Z> X g(Y);
void h() {
  int i = f<int>(5.6);          // Y is deduced to be double
  int j = f(5.6);               // ill-formed: X cannot be deduced
  f<void>(f<int, bool>);        // Y for outer f deduced to be int (*)(bool)
  f<void>(f<int>);              // ill-formed: f<int> does not denote a single function template specialization
  int k = g<int>(5.6);          // Y is deduced to be double, Z is deduced to an empty sequence
  f<void>(g<int, bool>);        // Y for outer f is deduced to be int (*)(bool),
                                // Z is deduced to an empty sequence

end example]

[Note: An empty template argument list can be used to indicate that a given use refers to a specialization of a function template even when a non-template function ([dcl.fct]) is visible that would otherwise be used. For example:

template <class T> int f(T);    // #1
int f(int);                     // #2
int k = f(1);                   // uses #2
int l = f<>(1);                 // uses #1

end note]

Template arguments that are present shall be specified in the declaration order of their corresponding template-parameters. The template argument list shall not specify more template-arguments than there are corresponding template-parameters unless one of the template-parameters is a template parameter pack. [Example:

template<class X, class Y, class Z> X f(Y,Z);
template<class ... Args> void f2();
void g() {
  f<int,const char*,double>("aa",3.0);
  f<int,const char*>("aa",3.0); // Z is deduced to be double
  f<int>("aa",3.0);             // Y is deduced to be const char*, and Z is deduced to be double
  f("aa",3.0);                  // error: X cannot be deduced
  f2<char, short, int, long>(); // OK

end example]

Implicit conversions (Clause [conv]) will be performed on a function argument to convert it to the type of the corresponding function parameter if the parameter type contains no template-parameters that participate in template argument deduction. [Note: Template parameters do not participate in template argument deduction if they are explicitly specified. For example,

template<class T> void f(T);

class Complex {

void g() {
  f<Complex>(1);    // OK, means f<Complex>(Complex(1))

end note]

[Note: Because the explicit template argument list follows the function template name, and because conversion member function templates and constructor member function templates are called without using a function name, there is no way to provide an explicit template argument list for these function templates. end note]

[Note: For simple function names, argument dependent lookup applies even when the function name is not visible within the scope of the call. This is because the call still has the syntactic form of a function call ([basic.lookup.unqual]). But when a function template with explicit template arguments is used, the call does not have the correct syntactic form unless there is a function template with that name visible at the point of the call. If no such name is visible, the call is not syntactically well-formed and argument-dependent lookup does not apply. If some such name is visible, argument dependent lookup applies and additional function templates may be found in other namespaces. [Example:

namespace A {
  struct B { };
  template<int X> void f(B);
namespace C {
  template<class T> void f(T t);
void g(A::B b) {
  f<3>(b);          // ill-formed: not a function call
  A::f<3>(b);       // well-formed
  C::f<3>(b);       // ill-formed; argument dependent lookup applies only to unqualified names
  using C::f;
  f<3>(b);          // well-formed because C​::​f is visible; then A​::​f is found by argument dependent lookup

end example] end note]

Template argument deduction can extend the sequence of template arguments corresponding to a template parameter pack, even when the sequence contains explicitly specified template arguments. [Example:

template<class ... Types> void f(Types ... values);

void g() {
  f<int*, float*>(0, 0, 0);     // Types is deduced to the sequence int*, float*, int

end example]