Question

我希望能够为成员函数声明和其他地方使用的指针使用单个C ++ typedef。如果我可以匹配非成员函数的结构，如下所示，那将是完美的：

#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;
}

我似乎无法避免非静态成员函数的类型准重复：

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);
}

我的实际代码无法真正使用＆＃34; auto f = &B::foo;＆＃34;无处不在，所以如果可能的话，我想尽量减少我的界面合同。是否有一个有效的语法来命名非指针成员函数类型？没有void(B::)(int)，void(B::&)(int)等的微不足道的变体。

编辑：接受的答案 - func_type Class::*语法是我所缺少的。谢谢，伙计们！

Answer 1

您似乎正在寻找的语法是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);
}

Above code at coliru.

C ++中的非指针成员函数typedef？

1 个答案: