printDad功能帮助 [英] printDad function help

问题描述

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推荐答案

在printDad函数中，替换

In printDad function, replace

成员8186414写道:

Member 8186414 wrote:

cout<< p.getAge(p.getDad())<< endl;

cout << p.getAge(p.getDad()) << endl;

与

with

cout << p.getDad()->getAge() << endl;

与问题不完全相关，但...
注意

Not strictly related to the question but ...
Be careful about

void printDad(Person p) 

实际上，它不是指您所指的人，而是指它的副本.
由于C ++默认副本构造函数是如何工作的，因此具有相同的成员值(因此指向相同的父母)，因此顺便说一遍也可以，但实际上是临时兄弟".
传递指针或引用，而不是值.

另外，学习使用const 表示不修改对象的成员函数，使用const&表示参数.您将避免大量无用的副本.

It actually doesn''t refer to the person you mean, but to a copy of it.
That -because of how C++ default copy constructor works- has the same members values (hence point to the same mom and dad) and hence incidentally will work the same, but it is actually a "temporary brother".
Pass a pointer or a reference, not a value.

Also, learn to use const for member function that don''t modify the object, and const& for parameters. You''ll avoid a lot useless copy.

void printDad(Person& p) // cause i cant see a copy constructor
{
  if(p.getDad())
    cout << p.getDad()->getAge() << endl;
  else
    cout << "orphan" << endl;
}

或实现副本构造函数:

or implement a copy constructor:

class Person
{
  Person* mom,
  *       dad;
  int     age;
public:
  Person(){ mom=dad=0; age=0; }
  Person(Person& init){ mom=init.mom; dad=init.dad; age=init.age; }
   // the others
};

祝你好运.

Good luck.

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