有一个我无法阻止编译的不良 C 风格转换。不受欢迎的强制转换执行 C 风格的强制转换，从某个类的对象到某个其他类的非常量引用。这些类(class)是无关的。同时，我喜欢支持从同一类的对象到 const 引用的 C 风格转换。我正在提供一个公共(public)转换运算符来支持理想的转换。在这种情况下，似乎无法阻止不受欢迎的转换。
转换为非常量引用无法构建(“Sandbox::B::operator Sandbox::A &()”(在第 30 行声明)不可访问*)，不幸的是转换为 const 引用要么失败(错误:不止一个从“Sandbox::B”到“const Sandbox::A”的转换函数适用: 函数“沙盒::B::operator const Sandbox::A &()” 函数“Sandbox::B::operator Sandbox::A &()”):

#include <iostream>
#include <string>
#include <cstdlib>

namespace Sandbox {
    class A {
    public:
        A (int i) : _x (i) { }
    private:
        int _x;
    };

    class B {
    public:
        B (const char* m) : _m (m), _a (std::atoi (m)) { }

        /*
         * This one shall be supported.
         */ 
        operator const A& () {
            return _a;
        }
    private:
        /*
         * This one shall be not supported.
         * If this one is disabled both desired and undesired conversions pass the compilation.
         */ 
        operator A& ();

        const std::string _m;
        const A _a;
    };
}

int main () {
    Sandbox::A a (1973);
    Sandbox::B b ("1984");

    /*
     * This is the undesirable cast and it shall fail to compile.
     */
    (Sandbox::A&)b;
    /*
     * This is the desirable cast and it shall pass the compilation.
     */
    (const Sandbox::A&)b;

    return 0;
}

如果我禁用运算符 operator A& () 所需和不需要的转换都会生成。

我正在使用 gcc、icc 和 MSVC 编译。 我无法控制客户端代码并阻止在那里使用 C 样式转换。

最佳答案

这应该可以解决问题(在 clang3.5 上测试过):

#include <iostream>
#include <string>
#include <cstdlib>

namespace Sandbox {
  class A {
  public:
    A (int i) : _x (i) { }

    void        fun()
    {
      std::cout << "action" << std::endl;
    }

  private:
    int _x;
  };

  class B {
  public:
    B (const char* m) : _m (m), _a (std::atoi (m)) { }

    /*
     * This one shall be supported.
     */
    template<typename T, typename Enable = typename std::enable_if<std::is_same<T, A>::value, A>::type>
    operator const T& ()
    {
      return _a;
    }

    /*
     * This one shall be not supported.
     * If this one is disabled both desired and undesired conversions pass the compilation.
     */
  private:
    template<typename T, typename Enable = typename std::enable_if<std::is_same<T, A>::value, A>::type>
    operator T& ();

    const std::string _m;
    const A _a;
  };
}

int main () {
  Sandbox::A a (1973);
  Sandbox::B b ("1984");

  /*
   * This is the undesirable cast and it shall fail to compile.
   */
  (Sandbox::A&)b;

  /*
   * This is the desirable cast and it shall pass the compilation.
   */
  (const Sandbox::A&)b;

  return 0;
}

至于为什么你的版本没有做到你想要的，跟C-Style cast的规则有关:

When the C-style cast expression is encountered, the compiler attempts the following cast expressions, in this order:

a) const_cast(expression)

b) static_cast(expression), with extensions: pointer or reference to a derived class is additionally allowed to be cast to pointer or reference to unambiguous base class (and vice versa) even if the base class is inaccessible (that is, this cast ignores the private inheritance specifier). Same applies to casting pointer to member to pointer to member of unambigous non-virtual base

c) static_cast (with extensions) followed by const_cast

d) reinterpret_cast(expression)

e) reinterpret_cast followed by const_cast

The first choice that satisfies the requirements of the respective cast operator is selected, even if it cannot be compiled

免责声明:这个解释主要基于猜测，有多个步骤和复杂的规则，所以我不确定一切是否真的有效，因为我认为我已经理解了，但你看吧。

由于您转换为引用，reinterpret_cast 将始终基于其 rules of type aliasing 工作，因此使 C 风格转换失败的唯一方法是对该类型进行 static_cast 明确地产生错误。不幸的是，转换规则似乎并不认为用户定义的转换为 const 类型比用户定义的转换为非 cv 限定类型更好，它们甚至处于同一级别如果 static_cast 目标类型是 const 限定的。鉴于使用模板、SFINAE 和参数推导，以及从山龙中提取的一些神奇编译器粉末，它起作用了。 (是的，这一步对我来说也有点神秘)。

关于c++ - 为什么我无法阻止不受欢迎的 C 风格强制转换进行编译？，我们在Stack Overflow上找到一个类似的问题： https://stackoverflow.com/questions/26463136/