### 12.4.4 Best viable function [over.match.best]

#### 12.4.4.1 General [over.match.best.general]

Define ICSi(F) as follows:
• If F is a static member function, ICS1(F) is defined such that ICS1(F) is neither better nor worse than ICS1(G) for any function G, and, symmetrically, ICS1(G) is neither better nor worse than ICS1(F);127 otherwise,
• let ICSi(F) denote the implicit conversion sequence that converts the i-th argument in the list to the type of the i-th parameter of viable function F.
[over.best.ics] defines the implicit conversion sequences and [over.ics.rank] defines what it means for one implicit conversion sequence to be a better conversion sequence or worse conversion sequence than another.
Given these definitions, a viable function F1 is defined to be a better function than another viable function F2 if for all arguments i, ICSi(F1) is not a worse conversion sequence than ICSi(F2), and then
• for some argument j, ICSj(F1) is a better conversion sequence than ICSj(F2), or, if not that,
• the context is an initialization by user-defined conversion (see [dcl.init], [over.match.conv], and [over.match.ref]) and the standard conversion sequence from the return type of F1 to the destination type (i.e., the type of the entity being initialized) is a better conversion sequence than the standard conversion sequence from the return type of F2 to the destination type
[Example 1: struct A { A(); operator int(); operator double(); } a; int i = a; // a.operator int() followed by no conversion is better than // a.operator double() followed by a conversion to int float x = a; // ambiguous: both possibilities require conversions, // and neither is better than the other — end example]
or, if not that,
• the context is an initialization by conversion function for direct reference binding of a reference to function type, the return type of F1 is the same kind of reference (lvalue or rvalue) as the reference being initialized, and the return type of F2 is not
[Example 2: template <class T> struct A { operator T&(); // #1 operator T&&(); // #2 }; typedef int Fn(); A<Fn> a; Fn& lf = a; // calls #1 Fn&& rf = a; // calls #2 — end example]
or, if not that,
• F1 is not a function template specialization and F2 is a function template specialization, or, if not that,
• F1 and F2 are function template specializations, and the function template for F1 is more specialized than the template for F2 according to the partial ordering rules described in [temp.func.order], or, if not that,
• F1 and F2 are non-template functions with the same parameter-type-lists, and F1 is more constrained than F2 according to the partial ordering of constraints described in [temp.constr.order], or if not that,
• F1 is a constructor for a class D, F2 is a constructor for a base class B of D, and for all arguments the corresponding parameters of F1 and F2 have the same type
[Example 3: struct A { A(int = 0); }; struct B: A { using A::A; B(); }; int main() { B b; // OK, B​::​B() } — end example]
or, if not that,
• F2 is a rewritten candidate ([over.match.oper]) and F1 is not
[Example 4: struct S { friend auto operator<=>(const S&, const S&) = default; // #1 friend bool operator<(const S&, const S&); // #2 }; bool b = S() < S(); // calls #2 — end example]
or, if not that,
• F1 and F2 are rewritten candidates, and F2 is a synthesized candidate with reversed order of parameters and F1 is not
[Example 5: struct S { friend std::weak_ordering operator<=>(const S&, int); // #1 friend std::weak_ordering operator<=>(int, const S&); // #2 }; bool b = 1 < S(); // calls #2 — end example]
or, if not that
• F1 is generated from a deduction-guide ([over.match.class.deduct]) and F2 is not, or, if not that,
• F1 is the copy deduction candidate and F2 is not, or, if not that,
• F1 is generated from a non-template constructor and F2 is generated from a constructor template.
[Example 6: template <class T> struct A { using value_type = T; A(value_type); // #1 A(const A&); // #2 A(T, T, int); // #3 template<class U> A(int, T, U); // #4 // #5 is the copy deduction candidate, A(A) }; A x(1, 2, 3); // uses #3, generated from a non-template constructor template <class T> A(T) -> A<T>; // #6, less specialized than #5 A a(42); // uses #6 to deduce A<int> and #1 to initialize A b = a; // uses #5 to deduce A<int> and #2 to initialize template <class T> A(A<T>) -> A<A<T>>; // #7, as specialized as #5 A b2 = a; // uses #7 to deduce A<A<int>> and #1 to initialize — end example]
If there is exactly one viable function that is a better function than all other viable functions, then it is the one selected by overload resolution; otherwise the call is ill-formed.128
[Example 7: void Fcn(const int*, short); void Fcn(int*, int); int i; short s = 0; void f() { Fcn(&i, s); // is ambiguous because &i int* is better than &i const int* // but s short is also better than s int Fcn(&i, 1L); // calls Fcn(int*, int), because &i int* is better than &i const int* // and 1L short and 1L int are indistinguishable Fcn(&i, 'c'); // calls Fcn(int*, int), because &i int* is better than &i const int* // and c int is better than c short } — end example]
If the best viable function resolves to a function for which multiple declarations were found, and if at least two of these declarations — or the declarations they refer to in the case of using-declarations — specify a default argument that made the function viable, the program is ill-formed.
[Example 8: namespace A { extern "C" void f(int = 5); } namespace B { extern "C" void f(int = 5); } using A::f; using B::f; void use() { f(3); // OK, default argument was not used for viability f(); // error: found default argument twice } — end example]
If a function is a static member function, this definition means that the first argument, the implied object argument, has no effect in the determination of whether the function is better or worse than any other function.

The algorithm for selecting the best viable function is linear in the number of viable functions.
Run a simple tournament to find a function W that is not worse than any opponent it faced.
Although another function F that W did not face might be at least as good as W, F cannot be the best function because at some point in the tournament F encountered another function G such that F was not better than G.
Hence, either W is the best function or there is no best function.
So, make a second pass over the viable functions to verify that W is better than all other functions.