C++ Primer 笔记--13章：定义值型类

要使指针成员表现得像一个值，复制 HasPtr 对象时必须复制指针所指向的对象。
复制构造函数不再复制指针，它将分配一个新的 int 对象，并初始化该对象以保存与被复制对象相同的值。每个对象都保存属于自己的 int 值的不同副本。因为每个对象保存自己的副本，所以析构函数将无条件删除指针。

赋值操作符不需要分配新对象，它只是必须记得给其指针所指向的对象赋新值，而不是给指针本身赋值。

/* * Valuelike behavior even though HasPtr has a pointer: * Each time we copy a HasPtr object, we make a new copy of the underlying int object to which ptr points */class HasPtr{public://no point to passing a pointer if we're going to copy it anyway//store pointer to a copy of the object we're givenHasPtr(const int &p, int i):ptr(new int(p)), val(i) {}//copy members and increment the use countHasPtr(const HasPtr &orig):ptr(new int(*orig.ptr)), val(orig.val) {}HasPtr& operator=(const HasPtr &);~HasPtr() { delete ptr; }//accessors must change to fetch value from Ptr objectint get_ptr_val() const { return *ptr; }int get_int() const { return val; }//change the appropriate data membervoid set_ptr(int *p) { ptr = p; }void set_int(int i) { val = i; }//return or change the value pointed to, so ok for const objectsint *get_ptr() const { return ptr; }void  set_ptr_val(int p) const { *ptr = p; }private:int *ptr;int val;};HasPtr& HasPtr::operator=(const HasPtr &rhs){//Note:Every HasPtr is guaranteed to point an actual int ; We know that ptr cannot be a zero pointer*ptr = *rhs.ptr;val = rhs.val;return *this;}