# 13 Templates [temp]

## 13.5 Template constraints [temp.constr]

### 13.5.1 General [temp.constr.general]

[Note 1:
Subclause [temp.constr] defines the meaning of constraints on template arguments.
The abstract syntax and satisfaction rules are defined in [temp.constr.constr].
Constraints are associated with declarations in [temp.constr.decl].
Declarations are partially ordered by their associated constraints ([temp.constr.order]).
— end note]

### 13.5.2 Constraints [temp.constr.constr]

#### 13.5.2.1 General [temp.constr.constr.general]

A constraint is a sequence of logical operations and operands that specifies requirements on template arguments.
The operands of a logical operation are constraints.
There are three different kinds of constraints:
In order for a constrained template to be instantiated ([temp.spec]), its associated constraints shall be satisfied as described in the following subclauses.
[Note 1:
Forming the name of a specialization of a class template, a variable template, or an alias template ([temp.names]) requires the satisfaction of its constraints.
Overload resolution requires the satisfaction of constraints on functions and function templates.
— end note]

#### 13.5.2.2 Logical operations [temp.constr.op]

There are two binary logical operations on constraints: conjunction and disjunction.
[Note 1:
These logical operations have no corresponding C++ syntax.
For the purpose of exposition, conjunction is spelled using the symbol  ∧  and disjunction is spelled using the symbol  ∨ .
The operands of these operations are called the left and right operands.
In the constraint A  ∧ B, A is the left operand, and B is the right operand.
— end note]
A conjunction is a constraint taking two operands.
To determine if a conjunction is satisfied, the satisfaction of the first operand is checked.
If that is not satisfied, the conjunction is not satisfied.
Otherwise, the conjunction is satisfied if and only if the second operand is satisfied.
A disjunction is a constraint taking two operands.
To determine if a disjunction is satisfied, the satisfaction of the first operand is checked.
If that is satisfied, the disjunction is satisfied.
Otherwise, the disjunction is satisfied if and only if the second operand is satisfied.
[Example 1: template<typename T> constexpr bool get_value() { return T::value; } template<typename T> requires (sizeof(T) > 1) && (get_value<T>()) void f(T); // has associated constraint sizeof(T) > 1  ∧  get_value<T>() void f(int); f('a'); // OK, calls f(int)
In the satisfaction of the associated constraints of f, the constraint sizeof(char) > 1 is not satisfied; the second operand is not checked for satisfaction.
— end example]
[Note 2:
A logical negation expression ([expr.unary.op]) is an atomic constraint; the negation operator is not treated as a logical operation on constraints.
As a result, distinct negation constraint-expressions that are equivalent under [temp.over.link] do not subsume one another under [temp.constr.order].
Furthermore, if substitution to determine whether an atomic constraint is satisfied ([temp.constr.atomic]) encounters a substitution failure, the constraint is not satisfied, regardless of the presence of a negation operator.
Here, requires (!sad<typename T​::​type>) requires that there is a nested type that is not sad, whereas requires (!sad_nested_type<T>) requires that there is no sad nested type.
— end example]
— end note]

#### 13.5.2.3 Atomic constraints [temp.constr.atomic]

An atomic constraint is formed from an expression E and a mapping from the template parameters that appear within E to template arguments that are formed via substitution during constraint normalization in the declaration of a constrained entity (and, therefore, can involve the unsubstituted template parameters of the constrained entity), called the parameter mapping ([temp.constr.decl]).
[Note 1:
Atomic constraints are formed by constraint normalization.
— end note]
Two atomic constraints, and , are identical if they are formed from the same appearance of the same expression and if, given a hypothetical template A whose template-parameter-list consists of template-parameters corresponding and equivalent ([temp.over.link]) to those mapped by the parameter mappings of the expression, a template-id naming A whose template-arguments are the targets of the parameter mapping of is the same ([temp.type]) as a template-id naming A whose template-arguments are the targets of the parameter mapping of .
[Note 2:
The comparison of parameter mappings of atomic constraints operates in a manner similar to that of declaration matching with alias template substitution ([temp.alias]).
[Example 1: template <unsigned N> constexpr bool Atomic = true; template <unsigned N> concept C = Atomic<N>; template <unsigned N> concept Add1 = C<N + 1>; template <unsigned N> concept AddOne = C<N + 1>; template <unsigned M> void f() requires Add1<2 * M>; template <unsigned M> int f() requires AddOne<2 * M> && true; int x = f<0>(); // OK, the atomic constraints from concept C in both fs are Atomic<N> // with mapping similar to template <unsigned N> struct WrapN; template <unsigned N> using Add1Ty = WrapN<N + 1>; template <unsigned N> using AddOneTy = WrapN<N + 1>; template <unsigned M> void g(Add1Ty<2 * M> *); template <unsigned M> void g(AddOneTy<2 * M> *); void h() { g<0>(nullptr); // OK, there is only one g } — end example]
As specified in [temp.over.link], if the validity or meaning of the program depends on whether two constructs are equivalent, and they are functionally equivalent but not equivalent, the program is ill-formed, no diagnostic required.
[Example 2: template <unsigned N> void f2() requires Add1<2 * N>; template <unsigned N> int f2() requires Add1<N * 2> && true; void h2() { f2<0>(); // ill-formed, no diagnostic required: // requires determination of subsumption between atomic constraints that are // functionally equivalent but not equivalent } — end example]
— end note]
To determine if an atomic constraint is satisfied, the parameter mapping and template arguments are first substituted into its expression.
If substitution results in an invalid type or expression, the constraint is not satisfied.
Otherwise, the lvalue-to-rvalue conversion is performed if necessary, and E shall be a constant expression of type bool.
The constraint is satisfied if and only if evaluation of E results in true.
If, at different points in the program, the satisfaction result is different for identical atomic constraints and template arguments, the program is ill-formed, no diagnostic required.
[Example 3: template<typename T> concept C = sizeof(T) == 4 && !true; // requires atomic constraints sizeof(T) == 4 and !true template<typename T> struct S { constexpr operator bool() const { return true; } }; template<typename T> requires (S<T>{}) void f(T); // #1 void f(int); // #2 void g() { f(0); // error: expression S<int>{} does not have type bool } // while checking satisfaction of deduced arguments of #1; // call is ill-formed even though #2 is a better match — end example]

### 13.5.3 Constrained declarations [temp.constr.decl]

A template declaration ([temp.pre]) or templated function declaration ([dcl.fct]) can be constrained by the use of a requires-clause.
This allows the specification of constraints for that declaration as an expression:
Constraints can also be associated with a declaration through the use of type-constraints in a template-parameter-list or parameter-type-list.
Each of these forms introduces additional constraint-expressions that are used to constrain the declaration.
A declaration's associated constraints are defined as follows:
The formation of the associated constraints establishes the order in which constraints are instantiated when checking for satisfaction ([temp.constr.constr]).
[Example 1: template<typename T> concept C = true; template<C T> void f1(T); template<typename T> requires C<T> void f2(T); template<typename T> void f3(T) requires C<T>;
The functions f1, f2, and f3 have the associated constraint C<T>.
template<typename T> concept C1 = true; template<typename T> concept C2 = sizeof(T) > 0; template<C1 T> void f4(T) requires C2<T>; template<typename T> requires C1<T> && C2<T> void f5(T);
The associated constraints of f4 and f5 are C1<T>  ∧  C2<T>.
template<C1 T> requires C2<T> void f6(); template<C2 T> requires C1<T> void f7();
The associated constraints of f6 are C1<T>  ∧  C2<T>, and those of f7 are C2<T>  ∧  C1<T>.
— end example]
When determining whether a given introduced constraint-expression of a declaration in an instantiated specialization of a templated class is equivalent ([temp.over.link]) to the corresponding constraint-expression of a declaration outside the class body, is instantiated.
If the instantiation results in an invalid expression, the constraint-expressions are not equivalent.
[Note 1:
This can happen when determining which member template is specialized by an explicit specialization declaration.
— end note]
[Example 2: template <class T> concept C = true; template <class T> struct A { template <class U> U f(U) requires C<typename T::type>; // #1 template <class U> U f(U) requires C<T>; // #2 }; template <> template <class U> U A<int>::f(U u) requires C<int> { return u; } // OK, specializes #2
Substituting int for T in C<typename T​::​type> produces an invalid expression, so the specialization does not match #1.
Substituting int for T in C<T> produces C<int>, which is equivalent to the constraint-expression for the specialization, so it does match #2.
— end example]

### 13.5.4 Constraint normalization [temp.constr.normal]

The normal form of an expression E is a constraint that is defined as follows:
• The normal form of an expression ( E ) is the normal form of E.
• The normal form of an expression E1 || E2 is the disjunction of the normal forms of E1 and E2.
• The normal form of an expression E1 && E2 is the conjunction of the normal forms of E1 and E2.
• The normal form of a concept-id C<A, A, ..., A> is the normal form of the constraint-expression of C, after substituting A, A, ..., A for C's respective template parameters in the parameter mappings in each atomic constraint.
If any such substitution results in an invalid type or expression, the program is ill-formed; no diagnostic is required.
[Example 1: template<typename T> concept A = T::value || true; template<typename U> concept B = A<U*>; template<typename V> concept C = B<V&>;
Normalization of B's constraint-expression is valid and results in T​::​value (with the mapping )  ∨  true (with an empty mapping), despite the expression T​::​value being ill-formed for a pointer type T.
Normalization of C's constraint-expression results in the program being ill-formed, because it would form the invalid type V&* in the parameter mapping.
— end example]
• The normal form of any other expression E is the atomic constraint whose expression is E and whose parameter mapping is the identity mapping.
The process of obtaining the normal form of a constraint-expression is called normalization.
[Note 1:
Normalization of constraint-expressions is performed when determining the associated constraints ([temp.constr.constr]) of a declaration and when evaluating the value of an id-expression that names a concept specialization ([expr.prim.id]).
— end note]
[Example 2: template<typename T> concept C1 = sizeof(T) == 1; template<typename T> concept C2 = C1<T> && 1 == 2; template<typename T> concept C3 = requires { typename T::type; }; template<typename T> concept C4 = requires (T x) { ++x; }; template<C2 U> void f1(U); // #1 template<C3 U> void f2(U); // #2 template<C4 U> void f3(U); // #3
The associated constraints of #1 are sizeof(T) == 1 (with mapping )  ∧  1 == 2.

The associated constraints of #2 are requires { typename T​::​type; } (with mapping ).

The associated constraints of #3 are requires (T x) { ++x; } (with mapping ).
— end example]

### 13.5.5 Partial ordering by constraints [temp.constr.order]

A constraint P subsumes a constraint Q if and only if, for every disjunctive clause in the disjunctive normal form118 of P, subsumes every conjunctive clause in the conjunctive normal form119 of Q, where
• a disjunctive clause subsumes a conjunctive clause if and only if there exists an atomic constraint in for which there exists an atomic constraint in such that subsumes , and
• an atomic constraint A subsumes another atomic constraint B if and only if A and B are identical using the rules described in [temp.constr.atomic].
[Example 1:
Let A and B be atomic constraints.
The constraint A  ∧ B subsumes A, but A does not subsume A  ∧ B.
The constraint A subsumes A  ∨ B, but A  ∨ B does not subsume A.
Also note that every constraint subsumes itself.
— end example]
[Note 1:
The subsumption relation defines a partial ordering on constraints.
This partial ordering is used to determine
— end note]
A declaration D1 is at least as constrained as a declaration D2 if
• D1 and D2 are both constrained declarations and D1's associated constraints subsume those of D2; or
• D2 has no associated constraints.
A declaration D1 is more constrained than another declaration D2 when D1 is at least as constrained as D2, and D2 is not at least as constrained as D1.
[Example 2: template<typename T> concept C1 = requires(T t) { --t; }; template<typename T> concept C2 = C1<T> && requires(T t) { *t; }; template<C1 T> void f(T); // #1 template<C2 T> void f(T); // #2 template<typename T> void g(T); // #3 template<C1 T> void g(T); // #4 f(0); // selects #1 f((int*)0); // selects #2 g(true); // selects #3 because C1<bool> is not satisfied g(0); // selects #4 — end example]
118)118)
A constraint is in disjunctive normal form when it is a disjunction of clauses where each clause is a conjunction of atomic constraints.
For atomic constraints A, B, and C, the disjunctive normal form of the constraint A  ∧ (B  ∨ C) is (A  ∧ B)  ∨ (A  ∧ C).
Its disjunctive clauses are (A  ∧ B) and (A  ∧ C).
119)119)
A constraint is in conjunctive normal form when it is a conjunction of clauses where each clause is a disjunction of atomic constraints.
For atomic constraints A, B, and C, the constraint A  ∧ (B  ∨ C) is in conjunctive normal form.
Its conjunctive clauses are A and (B  ∨ C).