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I am trying to use an method that takes
std::function<void()>
as an input. Unfortunately I have never used this data type before and I am having problems understanding it. All I need to do it say when it is called execute a method, for example:
std::function<void()> doThing = object.isMethod();
Although this syntax is obviously not correct. Can someone please explain this to me?
621
The stored callable object is called the target of std::function . If a std::function contains no target, it is called empty. Invoking the target of an empty std::function results in std::bad_function_call exception being thrown.
Instances of std::function can store, copy, and invoke any CopyConstructible Callable target -- functions, lambda expressions, bind expressions, or other function objects, as well as pointers to member functions and pointers to data members.
No. One is a function pointer; the other is an object that serves as a wrapper around a function pointer. They pretty much represent the same thing, but std::function is far more powerful, allowing you to do make bindings and whatnot.
When used for a function's parameter list, void specifies that the function takes no parameters. When used in the declaration of a pointer, void specifies that the pointer is "universal." If a pointer's type is void* , the pointer can point to any variable that's not declared with the const or volatile keyword.
You could bind the object to the method
std::function<void()> doThing = std::bind(&type::isMethod, &object);
or use a lambda
std::function<void()> doThing = [&]{object.isMethod();};
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