Is There a way to use the Parameter Names from a typedef

问题

So given a typedef that defines a function pointer with parameter names like this:

typedef void(*FOO)(const int arg);

Is there a way that I can just use this function pointer to define the signature of my function? Obviously this won't work, but I'd like to somehow use the typedef to specify a function signature with a corresponding type:

FOO foo {
    cout << arg << endl;
}

Again, I know this doesn't work, and is bad syntax. It will just give the error:

error: arg was not declared in this scope

回答1:

What you are trying to do will not work. FOO is an alias for void(*)(const int). so

FOO foo {
    cout << arg << endl;
}

becomes

void(*)(const int) foo {
    cout << arg << endl;
}

and that just doesn't work. What you can do though is define a macro that takes a name and use that to stamp out a function signature. That would look like

#define MAKE_FUNCTION(NAME) void NAME(const int arg)

MAKE_FUNCTION(foo){ std::cout << arg * 5 << "\n"; }

MAKE_FUNCTION(bar){ std::cout << arg * 10 << "\n"; }

int main()
{
    foo(1);
    bar(2);
}