相关疑难解决方法(0)

我今天去了一个面试,并得到了这个有趣的问题.

除了内存泄漏和事实上没有虚拟dtor,为什么这个代码会崩溃？

#include <iostream>

//besides the obvious mem leak, why does this code crash?

class Shape
{
public:
    virtual void draw() const = 0;
};

class Circle : public Shape
{
public:
    virtual void draw() const { }

    int radius;
};

class Rectangle : public Shape
{
public:
    virtual void draw() const { }

    int height;
    int width;
};

int main()
{
    Shape * shapes = new Rectangle[10];
    for (int i = 0; i < 10; ++i)
        shapes[i].draw();
}

考虑一对相关结构的以下声明.后代类不添加任何成员变量,唯一的成员函数是一个构造函数,它不会做任何事情,只是将其所有参数转发给基类的构造函数.

struct Base {
  Base(int a, char const* b):
    a(a), b(b)
  { }
  int a;
  char const* b;
};

struct Descendant: Base {
  Descendant(int a, char const* b):
    Base(a, b)
  { }
};

现在考虑使用这些类型的以下代码.函数foo期望接收一个数组Base.但是,main定义一个数组Descendant并将其传递给它foo.

void foo(Base* array, unsigned len)
{
  /* <access contents of array> */
}

int main()
{
  Descendant main_array[] = {
    Descendant(0, "zero"),
    Descendant(1, "one")
  };
  foo(main_array, 2);
  return 0;
}

是否定义了该程序的行为？答案是否取决于它的主体foo,例如它是写入数组还是只读取它？

如果sizeof(Derived)不相等 …

我在C++中有以下代码:

#include <iostream> 

class Number 
{ 
public: 
    virtual void foo(){std::cout << "Number foo\n";};
    Number (){ std::cout << "Number ctor" << std::endl;} 
    virtual ~Number(){ std::cout << "Number dtor" << std::endl;} 
}; 


class Complex : public Number 
{ 
public:
    virtual void foo(){std::cout << "Complex foo\n";};
    Complex (double r=0, double i=0) : _r (r), _i (i)
    { std::cout << "Complex ctor" << std::endl; };
    virtual ~Complex(){ std::cout << "Complex dtor" << std::endl;}
private: 
    double _r,_i;
};


int main()
{
    Number *numArr = new …