Question

我刚学习C ++，所以提供的任何帮助都会很棒。我试图在我的代码中重载二元运算符+：这是我的实现。

我使用的编译器是Mac OS X上的Xcode 6.1。

我收到一个编译错误，隐式删除了复制赋值运算符，我在main函数中使用了+运算符。

#include <iostream>
#include <string>

using namespace std;

class Person{

private:
string name;
int age;
double height;

public:
Person(string n="Haroon", int a=23, double h=72.1){
    this->name = n;
    this->age = a;
    this->height = h;
};

int getAge() const { return this->age; }
double getHeight() const { return this->height; }
string getName() const { return this->name; }

void setHeight(double x){
    this->height = x;
}

Person operator+ (const Person &rhs){

    string a = this->getName();
    string b = rhs.getName();

    string name = a+b;

    int c = max(this->getAge(),rhs.getAge());
    int d = max(this->getHeight(),rhs.getHeight());

    Person p(name,c,d);
    return p;
}

~Person(){};
};

int main(int argc, const char * argv[]) {

Person a("Abc",1,2);
Person b("Def",5,65);

Person n;
n = a+b;

return 0;
}

Answer 1

原因在于：

人n; //您正在调用默认构造函数。你会得到一个人（“Haroon”，23,72.1）

n = a + b; //您正在调用一个没有重载的赋值运算符

C ++有默认构造函数，转换构造函数和复制构造函数。

[~/testing/constructor]$ cat main.cpp

#include <iostream>

class A
{
public:

A()      : value(0) { std::cout << "Default " << value << std::endl; }
A(int a) : value(a) { std::cout << "Convert " << value << std::endl; }
A(const A& other) : value(other.value) { std::cout << "Copy    " << value << std::endl; }

private:
    int value;
};

int main()
{
    A x;
    A y(10);
    A z = y;
    A w(x);
}


[~/testing/constructor]$ ./run
Default 0
Convert 10
Copy    10
Copy    0

这里x默认构造，y是转换。请注意，A z = y;和A w(x);都是复制构造函数。它们完全一样!!但是您的n = a + b;是分配运算符。

使用初始化而不是构造函数中的赋值。 : value(0)是初始化。这是Scott Meyers“Effective C ++”一书中的成语。

如果您正在编写类的成员函数，则可以在C ++中省略this->。您使用的变量是类或函数局部变量或全局变量的成员变量。（如果名称有冲突，使用优先级本地＆gt;成员＆gt;全局变量）例如。

[~/testing/var]$ cat main.cpp
#include <iostream>

int a;

class A
{
public:
    A()      : a(0) {}
    A(int x) : a(x) {}
    ~A() {}

    void f() { int a = 9; std::cout << a << std::endl; }
private:
    int a;
};

int main()
{
    A x(5);
    x.f();
}
[~/testing/var]$ ./run
9

对于你的班级，我会写这样的正式C ++格式：

class Person{

public: // constructor & destructor
    Person(string n="Haroon", int a=23, double h=72.1)
        : name(n)
        , age(a)
        , height(h)
    {
    }
    Person(const Person& other)
        : name(other.name)
        , age(other.age)
        , height(other.height)
    {
    }

public: // operators

    Person& operator=(const Person& other)
    {
        if (&other == this) return *this;  // Such as using as "a = a"

        name   = other.name;
        age    = other.age;
        height = other.height;

        return *this;
    }

    Person operator+(const Person& other)
    {
        string n = name + other.name;  // You don't have to use other.getName() because C++ private variable is accessible by self class and friend class
        age    a = max(age, other.age);
        height h = max(height, other.height);

        return Person(n, a, h);
    }
public: // setter & getter

    int    getAge()    const { return age;    }
    double getHeight() const { return height; }
    string getName()   const { return name;   }

    void setHeight(double x) { height = x; }

private:
    string name;
    int    age;
    double height;

};

二进制+运算符重载c ++

1 个答案: