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Function taking any argument
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| #include <iostream> | |
| #include <functional> | |
| #include <any> | |
| struct any_type : std::any | |
| { | |
| using std::any::any; | |
| template <class T> operator T() const { | |
| return std::any_cast<T>(*this); | |
| } | |
| }; | |
| using function_one = | |
| std::function<void(any_type)>; | |
| int main() | |
| { | |
| function_one fn; | |
| fn = [](double x) { | |
| std::cout << x << '\n'; | |
| }; | |
| fn(3.14); | |
| fn = [](const char* str) { | |
| std::cout << str << '\n'; | |
| }; | |
| fn("test"); | |
| } |
Will raise std::bad_any_cast if type of call argument differ to argument type of lambda.
Actually this is an expanded version of std::function<void(std::any)> where std::any does not have a cast operator. The only acceptable argument is auto or std::any.
std::function<void(std::any)> fn;
fn = [](auto str) {
std::cout << std::any_cast<const char*>(str) << '\n';
};
fn("test");
With return type
using function_one = std::function<any_type(any_type)>;
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Output: