I'd like to be able to use a single C++ typedef for both member function declarations and for pointers to them used elsewhere. If I could match the structure of non-member functions like the following, it would be perfect:
#include <iostream>
using x_ft = char (int);
// Forward decls both advertise and enforce shared interface.
x_ft foo, bar;
char foo(int n) { return (n % 3 == 0) ? 'a' : 'b'; }
char bar(int n) { return (n % 5 == 0) ? 'c' : 'd'; }
int main(int argc, char *argv[]) {
// Using same typedef for pointers as the non-pointers above.
x_ft *f{foo};
x_ft *const g{(argc % 2 == 0) ? bar : foo};
std::cout << f(argc) << ", " << g(argc * 7) << std::endl;
}
I don't seem to be able to avoid type quasi-duplication for non-static member functions though:
struct B {
// The following works OK, despite vi_f not being class-member-specific.
using vi_ft = void(int);
vi_ft baz, qux;
// I don't want redundant definitions like these if possible.
using vi_pm_ft = void(B::*)(int); // 'decltype(&B::baz)' no better IMO.
};
void B::baz(int n) { /* ... */ }
void B::qux(int n) { /* ... */ }
void fred(bool x) {
B b;
B::vi_pm_ft f{&B::baz}; // A somehow modified B::vi_f would be preferable.
// SYNTAX FROM ACCEPTED ANSWER:
B::vi_ft B::*g{x ? &B::baz : &B::qux}; // vi_ft not needed in B:: now.
B::vi_ft B::*h{&B::baz}, B::*i{&B::qux};
(b.*f)(0);
(b.*g)(1);
(b.*h)(2);
(b.*i)(3);
}
My actual code can't really use sidesteps like "auto f = &B::foo;
" everywhere, so I'd like to minimize my interface contracts if at all possible. Is there a valid syntax for naming a non-pointer member function type? No trivial variants of void(B::)(int)
, void(B::&)(int)
, etc. work.
Edit: Accepted answer - the func_type Class::*
syntax is what I was missing. Thanks, guys!
The syntax you seem to be looking for is vi_ft B::*f
.
using vi_ft = void(int);
class B {
public:
vi_ft baz, qux;
};
void B::baz(int) {}
void B::qux(int) {}
void fred(bool x) {
B b;
vi_ft B::*f{ x ? &B::baz : &B::qux };
(b.*f)(0);
}
int main() {
fred(true);
fred(false);
}