带有继承和强制转换的段错误

Question

我正在使用一个现有的C++，看起来像这样

class Geometry {
public:
    enum {BOX, MESH} type;
    std::string name;
};

class Mesh : public Geometry {
public:
    std::string filename;
};

std::shared_ptr<Geometry> getGeometry() {
    std::shared_ptr<Geometry> geom = std::make_shared<Geometry>();
    geom->name = "geometry";

    Mesh *mesh = new Mesh();
    mesh->name = "name";
    mesh->filename = "filename";
    mesh->type = Geometry::MESH;
    geom.reset(mesh);

    return geom;
}

我需要扩展这些类提供的功能，所以我派生了这两个类

#include <memory>
#include <iostream>

class Geometry {
public:
    enum {BOX, MESH} type;
    std::string name;
};

class Mesh : public Geometry {
public:
    std::string filename;
};

std::shared_ptr<Geometry> getGeometry() {
    std::shared_ptr<Geometry> geom = std::make_shared<Geometry>();
    geom->name = "geometry";

    Mesh *mesh = new Mesh();
    mesh->name = "name";
    mesh->filename = "filename";
    mesh->type = Geometry::MESH;
    geom.reset(mesh);

    return geom;
}

class GeometryDerived : public Geometry {
public:
    void consumeGeometry() {
        Geometry geom = static_cast<Geometry>(*this);
        std::cout << geom.name << std::endl;
        if(geom.type == Geometry::MESH) {
            Mesh *mesh = static_cast<Mesh*>(&geom);
            std::cout << mesh->filename << std::endl;
            // mesh->get_output();
        }
    }
};

class MeshDerived : public Mesh {
public:
    std::string get_output() {
        return this->filename;        
    }
};



int main(int argc, char** argv)
{
    std::shared_ptr<Geometry> geom = getGeometry(); // this object is returned by the library

    std::shared_ptr<GeometryDerived> geomDerived = std::static_pointer_cast<GeometryDerived>(geom);
    geomDerived->consumeGeometry();

    return 0;
}

我得到一个segfault 行（在我的实际代码中，它的垃圾字符串）

std::cout << mesh->filename << std::endl;

虽然其他ints 和doubles 打印正确。即使geom.name 也能正确打印。这里的字符串是怎么回事？在这里投射物体的正确方法是什么？两次投射物体看起来很难看。下面是GDB 的输出

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Reading symbols from ./cpp-example...
(gdb) run
Starting program: /home/user/cpp-example
name

Program received signal SIGSEGV, Segmentation fault.
__memmove_avx_unaligned_erms () at ../sysdeps/x86_64/multiarch/memmove-vec-unaligned-erms.S:383
383 ../sysdeps/x86_64/multiarch/memmove-vec-unaligned-erms.S: No such file or directory.
(gdb) bt
#0  __memmove_avx_unaligned_erms () at ../sysdeps/x86_64/multiarch/memmove-vec-unaligned-erms.S:383
#1  0x00007ffff7c247b2 in _IO_new_file_xsputn (n=140737488346512, data=0xdb8bbc853a3b3400, f=<optimized out>) at fileops.c:1236
#2  _IO_new_file_xsputn (f=0x7ffff7d7e6a0 <_IO_2_1_stdout_>, data=0xdb8bbc853a3b3400, n=140737488346512) at fileops.c:1197
#3  0x00007ffff7c18541 in __GI__IO_fwrite (buf=0xdb8bbc853a3b3400, size=1, count=140737488346512, fp=0x7ffff7d7e6a0 <_IO_2_1_stdout_>) at libioP.h:948
#4  0x00007ffff7ed2824 in std::basic_ostream<char, std::char_traits<char> >& std::__ostream_insert<char, std::char_traits<char> >(std::basic_ostream<char, std::char_traits<char> >&, char const*, long) ()
   from /lib/x86_64-linux-gnu/libstdc++.so.6
#5  0x0000555555556858 in GeometryDerived::consumeGeometry (this=0x55555556def0) at ../downcasting.cpp:23
#6  0x0000555555556513 in main (argc=1, argv=0x7fffffffdec8) at ../downcasting.cpp:53

编辑 1： 以下更改工作正常，但这不是我想要的

int main(int argc, char** argv)
{
    std::shared_ptr<Geometry> geom = getGeometry();

    if(geom->type == Geometry::MESH) {
        std::shared_ptr<MeshDerived> meshDerived = std::static_pointer_cast<MeshDerived>(geom);
        std::cout << meshDerived->filename << std::endl;
    }

    // std::shared_ptr<GeometryDerived> geomDerived = std::static_pointer_cast<GeometryDerived>(geom);
    // geomDerived->consumeGeometry();

    return 0;
}

Answer 1

您的代码存在多个问题：

• 在GeometryDerived::consumeGeometry中，Geometry geom = static_cast<Geometry>(*this);通过slicing当前对象创建一个Geometry类型的新对象。稍后在该函数中，您执行Mesh *mesh = static_cast<Mesh*>(&geom);，它返回一个无效指针（因为geom 不是Mesh）。通过该指针访问数据是 UB。 （由于大多数编译器使用的内存布局，它可能适用于大多数实现中在 Geometry 中声明的数据成员；这并不能很好地定义行为。）

• static_pointer_cast 不检查指向的对象是否实际上是请求的类型。即使对象是不同的类型，转换也会成功，但返回的指针将是无效的。 getGeometry 函数返回一个指向 Mesh 对象的指针，该对象与 GeometryDerived 无关，因此强制转换返回一个无效指针。

---

对 Edit 1 部分的评论。不，这行也是错误的：

std::shared_ptr<MeshDerived> meshDerived = std::static_pointer_cast<MeshDerived>(geom);

原因和我上面提到的一样：getGeometry 返回一个指向Mesh 的指针，而不是MeshDerived。该代码可能适用于大多数编译器，因为 MeshDerived 与 Mesh 具有相同的内存布局，但这仍然不是有效的 C++。

您似乎想在运行时向现有类“添加方法”。这在 C++ 中是不可能的。您必须在主类中定义所需的方法（例如Mesh）或使用visitor pattern，即在一个单独的类中定义新功能，该类将通过其公共成员对Mesh等进行操作。