从书籍中复制的代码无法编译。已弃用或错别字?

Code copied from a book won't compile. Deprecated or typo?

本文关键字:错别字 编译 复制 代码      更新时间:2023-10-16

这是一个来自免费c++电子书的程序,但它无法编译。一开始我把它打出来,但意识到它无法编译,我试着直接从电子书中复制到代码块13.12中。它仍然无法编译。这本电子书是2010年出版的,所以可能代码不符合当前标准,或者在某个地方有语法错别字。请帮我找出问题所在。

错误是:

error: extra qualification 'Critter::' on member 'operator=' [-fpermissive]|

代码是:

#include <iostream>
using namespace std;

class Critter
{
public:
    Critter(const string& name = "", int age = 0);
    ~Critter(); //destructor prototype
    Critter(const Critter& c); //copy constructor prototype
    Critter& Critter::operator=(const Critter& c); 
op
    void Greet() const;
private:
    string* m_pName;
    int m_Age;
};
Critter::Critter(const string& name, int age)
{
    cout << "Constructor calledn";
    m_pName = new string(name);
    m_Age = age;
}
Critter::~Critter() //destructor definition
{
    cout << "Destructor calledn";
    delete m_pName;
}
Critter::Critter(const Critter& c) //copy constructor definition
{
    cout << "Copy Constructor calledn";
    m_pName = new string(*(c.m_pName));
    m_Age = c.m_Age;
}
Critter& Critter::operator=(const Critter& c) 
{
    cout << "Overloaded Assignment Operator calledn";
    if (this != &c)
    {
        delete m_pName;
        m_pName = new string(*(c.m_pName));
        m_Age = c.m_Age;
    }
    return *this;
}
void Critter::Greet() const
{
    cout << "I’m " << *m_pName << " and I’m " << m_Age << " years old.n";
    cout << "&m_pName: " << cout << &m_pName << endl;
}
void testDestructor();
void testCopyConstructor(Critter aCopy);
void testAssignmentOp();
int main()
{
    testDestructor();
    cout << endl;
    Critter crit("Poochie", 5);
    crit.Greet();
    testCopyConstructor(crit);
    crit.Greet();
    cout << endl;
    testAssignmentOp();
    return 0;
}
void testDestructor()
{
    Critter toDestroy("Rover", 3);
    toDestroy.Greet();
}
void testCopyConstructor(Critter aCopy)
{
    aCopy.Greet();
}
void testAssignmentOp()
{
    Critter crit1("crit1", 7);
    Critter crit2("crit2", 9);
    crit1 = crit2;
    crit1.Greet();
    crit2.Greet();
    cout << endl;
    Critter crit3("crit", 11);
    crit3 = crit3;
    crit3.Greet();
}

正如消息所说,

中有一个额外的Critter::
Critter& Critter::operator=(const Critter& c); 

由于这是在类声明中,所以它必须是类的成员,不需要以类名作为前缀。

Critter& operator=(const Critter& c); 

是正确的形式。

类名前缀仅在类之外定义成员时使用,如示例后面的代码。

从操作符原型中删除Critter::