A program that uses auto or decltype(auto) in a context not explicitly allowed in this section is ill-formed.
If the init-declarator-list contains more than one init-declarator, they shall all form declarations of variables. The type of each declared variable is determined by placeholder type deduction, and if the type that replaces the placeholder type is not the same in each deduction, the program is ill-formed.
[ Example:
auto x = 5, *y = &x; auto a = 5, b = { 1, 2 };
— end example ]
If a function with a declared return type that contains a placeholder type has multiple non-discarded return statements, the return type is deduced for each such return statement. If the type deduced is not the same in each deduction, the program is ill-formed.
If a function with a declared return type that uses a placeholder type has no non-discarded return statements, the return type is deduced as though from a return statement with no operand at the closing brace of the function body. [ Example:
auto f() { } auto* g() { }
— end example ]
If the type of an entity with an undeduced placeholder type is needed to determine the type of an expression, the program is ill-formed. Once a non-discarded return statement has been seen in a function, however, the return type deduced from that statement can be used in the rest of the function, including in other return statements. [ Example:
auto n = n; auto f();
void g() { &f; } auto sum(int i) {
if (i == 1)
return i; else
return sum(i-1)+i; }
— end example ]
Return type deduction for a function template with a placeholder in its declared type occurs when the definition is instantiated even if the function body contains a return statement with a non-type-dependent operand. [ Note: Therefore, any use of a specialization of the function template will cause an implicit instantiation. Any errors that arise from this instantiation are not in the immediate context of the function type and can result in the program being ill-formed ([temp.deduct]). — end note ] [ Example:
template <class T> auto f(T t) { return t; } typedef decltype(f(1)) fint_t; template<class T> auto f(T* t) { return *t; }
void g() { int (*p)(int*) = &f; }
— end example ]
Redeclarations or specializations of a function or function template with a declared return type that uses a placeholder type shall also use that placeholder, not a deduced type. [ Example:
auto f();
auto f() { return 42; } auto f(); int f(); decltype(auto) f();
template <typename T> auto g(T t) { return t; } template auto g(int); template char g(char); template<> auto g(double);
template <class T> T g(T t) { return t; } template char g(char); template auto g(float);
void h() { return g(42); }
template <typename T> struct A {
friend T frf(T);
};
auto frf(int i) { return i; }
— end example ]
A function declared with a return type that uses a placeholder type shall not be virtual.
An explicit instantiation declaration does not cause the instantiation of an entity declared using a placeholder type, but it also does not prevent that entity from being instantiated as needed to determine its type. [ Example:
template <typename T> auto f(T t) { return t; }
extern template auto f(int); int (*p)(int) = f;
— end example ]
10.1.7.4.1 Placeholder type deduction [dcl.type.auto.deduct]A type T containing a placeholder type, and a corresponding initializer e, are determined as follows:
for a non-discarded return statement that occurs in a function declared with a return type that contains a placeholder type, T is the declared return type and e is the operand of the return statement. If the return statement has no operand, then e is void();
for a variable declared with a type that contains a placeholder type, T is the declared type of the variable and e is the initializer. If the initialization is direct-list-initialization, the initializer shall be a braced-init-list containing only a single assignment-expression and e is the assignment-expression;
for a non-type template parameter declared with a type that contains a placeholder type, T is the declared type of the non-type template parameter and e is the corresponding template argument.
In the case of a return statement with no operand or with an operand of type void, T shall be either decltype(auto) or cv auto.
If the placeholder is the auto type-specifier, the deduced type T' replacing T is determined using the rules for template argument deduction. Obtain P from T by replacing the occurrences of auto with either a new invented type template parameter U or, if the initialization is copy-list-initialization, with std::initializer_list<U>. Deduce a value for U using the rules of template argument deduction from a function call, where P is a function template parameter type and the corresponding argument is e. If the deduction fails, the declaration is ill-formed. Otherwise, T' is obtained by substituting the deduced U into P. [ Example:
auto x1 = { 1, 2 }; auto x2 = { 1, 2.0 }; auto x3{ 1, 2 }; auto x4 = { 3 }; auto x5{ 3 };
— end example ]
[ Example:
const auto &i = expr;
The type of i is the deduced type of the parameter u in the call f(expr) of the following invented function template:
template <class U> void f(const U& u);
— end example ]
If the placeholder is the decltype(auto) type-specifier, T shall be the placeholder alone. The type deduced for T is determined as described in [dcl.type.simple], as though e had been the operand of the decltype. [ Example:
int i;
int&& f();
auto x2a(i); decltype(auto) x2d(i); auto x3a = i; decltype(auto) x3d = i; auto x4a = (i); decltype(auto) x4d = (i); auto x5a = f(); decltype(auto) x5d = f(); auto x6a = { 1, 2 }; decltype(auto) x6d = { 1, 2 }; auto *x7a = &i; decltype(auto)*x7d = &i;
— end example ]
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