Wrap overloaded function via std::function

问题

I have an overloaded function which I want to pass along wrapped in a std::function. GCC4.6 does not find a "matching function". While I did find some questions here the answers are not as clear as I would like them. Could someone tell me why the following code can not deduct the correct overload and how to (elegantly) work around it?

int test(const std::string&) {
    return 0;
}

int test(const std::string*) {
    return 0;
}

int main() {
    std::function<int(const std::string&)> func = test;
    return func();
}

回答1:

That is ambiguous situation.

To disambiguate it, use explicit cast as:

typedef int (*funtype)(const std::string&);

std::function<int(const std::string&)> func=static_cast<funtype>(test);//cast!

Now the compiler would be able to disambiguate the situation, based on the type in the cast.

Or, you can do this:

typedef int (*funtype)(const std::string&);

funtype fun = test; //no cast required now!
std::function<int(const std::string&)> func = fun; //no cast!

So why std::function<int(const std::string&)> does not work the way funtype fun = test works above?

Well the answer is, because std::function can be initialized with any object, as its constructor is templatized which is independent of the template argument you passed to std::function.

来源：https://stackoverflow.com/questions/10111042/wrap-overloaded-function-via-stdfunction