A function template defines an unbounded set of related functions. [ Example: a family of sort functions might be declared like this:
template<class T> class Array { }; template<class T> void sort(Array<T>&);
— end example ]
A function template can be overloaded with other function templates and with non-template functions ([dcl.fct]). A non-template function is not related to a function template (i.e., it is never considered to be a specialization), even if it has the same name and type as a potentially generated function template specialization.141
That is, declarations of non-template functions do not merely guide overload resolution of function template specializations with the same name. If such a non-template function is odr-used ([basic.def.odr]) in a program, it must be defined; it will not be implicitly instantiated using the function template definition.
It is possible to overload function templates so that two different function template specializations have the same type. [ Example:
// file1.c template<class T> void f(T*); void g(int* p) { f(p); // calls f<int>(int*) }
// file2.c template<class T> void f(T); void h(int* p) { f(p); // calls f<int*>(int*) }
— end example ]
Such specializations are distinct functions and do not violate the one definition rule ([basic.def.odr]).
The signature of a function template is defined in [intro.defs]. The names of the template parameters are significant only for establishing the relationship between the template parameters and the rest of the signature. [ Note: Two distinct function templates may have identical function return types and function parameter lists, even if overload resolution alone cannot distinguish them.
template<class T> void f(); template<int I> void f(); // OK: overloads the first template // distinguishable with an explicit template argument list
— end note ]
When an expression that references a template parameter is used in the function parameter list or the return type in the declaration of a function template, the expression that references the template parameter is part of the signature of the function template. This is necessary to permit a declaration of a function template in one translation unit to be linked with another declaration of the function template in another translation unit and, conversely, to ensure that function templates that are intended to be distinct are not linked with one another. [ Example:
template <int I, int J> A<I+J> f(A<I>, A<J>); // #1 template <int K, int L> A<K+L> f(A<K>, A<L>); // same as #1 template <int I, int J> A<I-J> f(A<I>, A<J>); // different from #1
— end example ] [ Note: Most expressions that use template parameters use non-type template parameters, but it is possible for an expression to reference a type parameter. For example, a template type parameter can be used in the sizeof operator. — end note ]
Two expressions involving template parameters are considered equivalent if two function definitions containing the expressions would satisfy the one definition rule ([basic.def.odr]), except that the tokens used to name the template parameters may differ as long as a token used to name a template parameter in one expression is replaced by another token that names the same template parameter in the other expression. For determining whether two dependent names ([temp.dep]) are equivalent, only the name itself is considered, not the result of name lookup in the context of the template. If multiple declarations of the same function template differ in the result of this name lookup, the result for the first declaration is used. [ Example:
template <int I, int J> void f(A<I+J>); // #1 template <int K, int L> void f(A<K+L>); // same as #1 template <class T> decltype(g(T())) h(); int g(int); template <class T> decltype(g(T())) h() // redeclaration of h() uses the earlier lookup { return g(T()); } // ...although the lookup here does find g(int) int i = h<int>(); // template argument substitution fails; g(int) // was not in scope at the first declaration of h()
— end example ] Two expressions involving template parameters that are not equivalent are functionally equivalent if, for any given set of template arguments, the evaluation of the expression results in the same value.
Two function templates are equivalent if they are declared in the same scope, have the same name, have identical template parameter lists, and have return types and parameter lists that are equivalent using the rules described above to compare expressions involving template parameters. Two function templates are functionally equivalent if they are equivalent except that one or more expressions that involve template parameters in the return types and parameter lists are functionally equivalent using the rules described above to compare expressions involving template parameters. If a program contains declarations of function templates that are functionally equivalent but not equivalent, the program is ill-formed; no diagnostic is required.
[ Note: This rule guarantees that equivalent declarations will be linked with one another, while not requiring implementations to use heroic efforts to guarantee that functionally equivalent declarations will be treated as distinct. For example, the last two declarations are functionally equivalent and would cause a program to be ill-formed:
// Guaranteed to be the same template <int I> void f(A<I>, A<I+10>); template <int I> void f(A<I>, A<I+10>); // Guaranteed to be different template <int I> void f(A<I>, A<I+10>); template <int I> void f(A<I>, A<I+11>); // Ill-formed, no diagnostic required template <int I> void f(A<I>, A<I+10>); template <int I> void f(A<I>, A<I+1+2+3+4>);
— end note ]
If a function template is overloaded, the use of a function template specialization might be ambiguous because template argument deduction ([temp.deduct]) may associate the function template specialization with more than one function template declaration. Partial ordering of overloaded function template declarations is used in the following contexts to select the function template to which a function template specialization refers:
during overload resolution for a call to a function template specialization ([over.match.best]);
when the address of a function template specialization is taken;
when a placement operator delete that is a function template specialization is selected to match a placement operator new ([basic.stc.dynamic.deallocation], [expr.new]);
when a friend function declaration ([temp.friend]), an explicit instantiation ([temp.explicit]) or an explicit specialization ([temp.expl.spec]) refers to a function template specialization.
Partial ordering selects which of two function templates is more specialized than the other by transforming each template in turn (see next paragraph) and performing template argument deduction using the function type. The deduction process determines whether one of the templates is more specialized than the other. If so, the more specialized template is the one chosen by the partial ordering process.
To produce the transformed template, for each type, non-type, or template template parameter (including template parameter packs ([temp.variadic]) thereof) synthesize a unique type, value, or class template respectively and substitute it for each occurrence of that parameter in the function type of the template. If only one of the function templates is a non-static member of some class A, that function template is considered to have a new first parameter inserted in its function parameter list. Given cv as the cv-qualifiers of the function template (if any), the new parameter is of type “rvalue reference to cv A” if the optional ref-qualifier of the function template is &&, or of type “lvalue reference to cv A” otherwise. [ Note: This allows a non-static member to be ordered with respect to a nonmember function and for the results to be equivalent to the ordering of two equivalent nonmembers. — end note ] [ Example:
struct A { }; template<class T> struct B { template<class R> int operator*(R&); // #1 }; template<class T, class R> int operator*(T&, R&); // #2 // The declaration of B::operator* is transformed into the equivalent of // template<class R> int operator*(B<A>&, R&); // #1a int main() { A a; B<A> b; b * a; // calls #1a }
— end example ]
Using the transformed function template's function type, perform type deduction against the other template as described in [temp.deduct.partial].
[ Example:
template<class T> struct A { A(); }; template<class T> void f(T); template<class T> void f(T*); template<class T> void f(const T*); template<class T> void g(T); template<class T> void g(T&); template<class T> void h(const T&); template<class T> void h(A<T>&); void m() { const int* p; f(p); // f(const T*) is more specialized than f(T) or f(T*) float x; g(x); // Ambiguous: g(T) or g(T&) A<int> z; h(z); // overload resolution selects h(A<T>&) const A<int> z2; h(z2); // h(const T&) is called because h(A<T>&) is not callable }
— end example ]
[ Note: Since partial ordering in a call context considers only parameters for which there are explicit call arguments, some parameters are ignored (namely, function parameter packs, parameters with default arguments, and ellipsis parameters). [ Example:
template<class T> void f(T); // #1 template<class T> void f(T*, int=1); // #2 template<class T> void g(T); // #3 template<class T> void g(T*, ...); // #4
int main() { int* ip; f(ip); // calls #2 g(ip); // calls #4 }
— end example ][ Example:
template<class T, class U> struct A { }; template<class T, class U> void f(U, A<U, T>* p = 0); // #1 template< class U> void f(U, A<U, U>* p = 0); // #2 template<class T > void g(T, T = T()); // #3 template<class T, class... U> void g(T, U ...); // #4 void h() { f<int>(42, (A<int, int>*)0); // calls #2 f<int>(42); // error: ambiguous g(42); // error: ambiguous }
— end example ][ Example:
template<class T, class... U> void f(T, U...); // #1 template<class T > void f(T); // #2 template<class T, class... U> void g(T*, U...); // #3 template<class T > void g(T); // #4 void h(int i) { f(&i); // error: ambiguous g(&i); // OK: calls #3 }
— end example ] — end note ]