C++ Primer 5th Edition: Pointer to member function

Question

Hello I have this text from C++ Primer 5th edition:

function<bool (const string&)> fcn = &string::empty; find_if(svec.begin(), svec.end(), fcn);

Here we tell function that empty is a function that can be called with a string and returns a bool. Ordinarily, the object on which a member function executes is passed to the implicit this parameter. When we want to use function to generate a callable for a member function, we have to “translate” the code to make that implicit parameter explicit.

So what he meant with: "When we want to use function... make that implicit parameter explicit"?

463035818_is_not_an_ai · Accepted Answer · 2021-08-02 20:45:02Z

It refers to the implicit this parameter to member functions. They get a pointer to the current object passed under the hood. std::function has some magic to turn that implicit parameter into an explicit one:

#include <iostream> #include <functional> struct foo { void bar() { std::cout << "Hello World\n";} }; int main() { std::function< void (foo&)> g = &foo::bar; foo f; f.bar(); // bar takes no parameters, but implicitly it gets a pointer to f g(f); // g(f) explicitly gets the parameter }

With f.bar() its the method call syntax that tells us that we call bar on the object f. f can be said to be an implicit parameter to bar. With g(f) that parameter is passed explicitly.

PS: Of course it isn't "magic", but I understood the question is about the general meaning of the implicit parameter, while explaining how std::function turns member functions into free callables is perhaps a topic for a different question.

Thanks a lot! You made it clear for me through such useul and easy explanation.

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