deduction guides for std::function - cppreference.com (original) (raw)

1. This overload participates in overload resolution only if &F::operator() is well-formed when treated as an unevaluated operand and decltype(&F::operator()) is of the form R(G::*)(A...) (optionally cv-qualified, optionally noexcept, optionally lvalue reference qualified). The deduced type is std::function<R(A...)>.

2. This overload participates in overload resolution only if &F::operator() is well-formed when treated as an unevaluated operand and F::operator() is an explicit object parameter function whose type is of form R(G, A...) or R(G, A...) noexcept. The deduced type is std::function<R(A...)>.

3. This overload participates in overload resolution only if &F::operator() is well-formed when treated as an unevaluated operand and F::operator() is a static member function whose type is of form R(A...) or R(A...) noexcept. The deduced type is std::function<R(A...)>.

[edit] Notes

These deduction guides do not allow deduction from a function with ellipsis parameter, and the ... in the types is always treated as a pack expansion.

The type deduced by these deduction guides may change in a later standard revision (in particular, this might happen if noexcept support is added to std::function in a later standard).

[edit] Example

#include int func(double) { return 0; } int main() { std::function f{func}; // guide #1 deduces function<int(double)> int i = 5; std::function g = & { return i; }; // guide #2 deduces function<int(double)> }

[edit] Defect reports

The following behavior-changing defect reports were applied retroactively to previously published C++ standards.