cppyy inherit class 包含一个智能指针

cppyy inherit class that contains a smart pointer

这是一个继承自包含智能指针的 class 的简单示例。我们不对它做任何事情,只是声明它。

import cppyy

cppyy.cppdef("""
  class Example { 
   private:
    std::unique_ptr<double> x;
   public:
    Example() {}
    virtual ~Example() = default;
    double y = 66.;
   };
  """)

class Inherit(cppyy.gbl.Example):
    pass

 a = Inherit()
 print(a.y)  # Test whether this attribute was inherited

示例运行,但出现智能指针错误

input_line_19:9:43: error: call to implicitly-deleted copy constructor of '::Example'
  Dispatcher1(const Dispatcher1& other) : Example(other), m_self(other.m_self, this) {}
                                          ^       ~~~~~
input_line_17:4:29: note: copy constructor of 'Example' is implicitly deleted because field 'x' has a deleted copy constructor
    std::unique_ptr<double> x;
                            ^
/usr/include/c++/7/bits/unique_ptr.h:383:7: note: 'unique_ptr' has been explicitly marked deleted here
      unique_ptr(const unique_ptr&) = delete;
      ^
smart_ptr.py:14: RuntimeWarning: no python-side overrides supported
  class Inherit(cppyy.gbl.Example):
66.0

尽管如此,继承似乎起作用了,因为我们仍然可以从 C++ class 访问 public 变量。实际上,我不是 100% 确定 cppyy 是否有问题。虽然 C++ 看起来不错,但我可能以一种奇怪的方式使用智能 pointers/the 虚拟析构函数,因为我对智能指针的使用经验不多。

如果我使用 std::shared_ptr 而不是 std::unique_ptr

,则不会引发错误

正如 S.M 所暗示的那样,如果我们必须使用 unique_ptr,诀窍似乎是确保定义一个复制构造函数,例如,这个例子给出了预期的结果而没有错误消息,

import cppyy

cppyy.cppdef("""
  class Example { 
    std::unique_ptr<double> x;
   public:
    Example() { x = std::unique_ptr<double>(new double(123.)); } 
    // Copy constructor
    Example(const Example& other) : x(other.x ? nullptr : new double(*other.x)) {}
    virtual ~Example() = default;
    double y = 66.;
    double get_x() { return *x; }
  };
  auto e = Example();
  auto f = e;
  """)

class Inherit(cppyy.gbl.Example):
  pass

a = Inherit()
print(a.get_x())  # prints 123.
print(a.y)  # prints 66.