std::function
Defined in header <functional> | ||
|---|---|---|
template<class R, class... ArgTypes> function(R(*)(ArgTypes...)) -> function<R(ArgTypes...)>; | (1) | (since C++17) |
template<class F> function(F) -> function</*see below*/>; | (2) | (since C++17) |
decltype(&F::operator()) is of the form R(G::*)(A...) (optionally cv-qualified, optionally noexcept, optionally lvalue reference qualified) for some class type G, then the deduced type is std::function<R(A...)>. This overload only participates in overload resolution if &F::operator() is well-formed when treated as an unevaluated operand. 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).
#include <functional>
int func(double) { return 0; }
int main() {
std::function f{func}; // guide #1 deduces function<int(double)>
int i = 5;
std::function g = [&](double) { return i; }; // guide #2 deduces function<int(double)>
}
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