• C++ 运算符重载


    14. 运算符重载

    14.1 Overloaded operator

    • Just a function with an operator name!
      • Use the operator keyword as a prefix to name operator *(...)
    • Can be a member function
      • Implicit first argument this:
        const String String::operator +(const String& that);
    • Can be a global (free) function
      • Both arguments explicit
        const String operator+(const String& r, const String& l);

    14.2 参数传入

    • If it is read-only, pass it in as a const reference.
    • Make member functions const that don't change the class(boolean operators, +, -, etc)
    • For global functions, if the left-hand side changes pass as a reference(assignment operators)

    14.3 结果返回

    • Select the return type depending on the expected meaning of the operator. For example,
      • For operator+ you need to generate a new object. Return as a const object so the result cannot be modified as an lvalue.
      • Logical operators should return bool(or int for older compilers)

    14.4 The prototypes of operators

    • +-*/%^&|~
      • const T operatorX(const T& left, const T& right) const;
    • !&&||< <= == >= >
      • bool operatorX(const T& left, const T& right) const;
    • []
      • T& T::operator[](int index);
    • operators ++ and --
    • Postfix forms take an int argument -- compiler will pass in 0 as that int
    class Integer {
    public:
        const Integer& operator++();    // prefix++
        const Integer operator++(int);  // postfix++
        const Integer& operator--();    // prefix--
        const Integer operator--(int);  // postfix--
    };
    
    const Integer& Integer::operator++() {
        *this += 1;
        return *this;
    }
    
    const Integer Integer::operator++(int) {
        Integer old(*this);
        ++(*this);
        return old;
    }
    
    • Relation operators
      • implement != in terms of ==
      • implement >, >=, <= in terms of <
    class Integer {
    public:
        bool operator==(const Integer& rhs) const;
        bool operator!=(const Integer& rhs) const;
        bool operator<(const Integer& rhs) const;
        bool operator>(const Integer& rhs) const;
        bool operator<=(const Integer& rhs) const;
        bool operator>=(const Integer& rhs) const;
    }
    
    bool Integer::operator==(const Integer& rhs) const {
        return i == rhs.i;
    }
    
    bool Integer::operator!=(const Integer& rhs) const {
        return !(*this == rhs);
    }
    
    bool Integer::operator<(const Integer& rhs) const {
        return i < rhs.i;
    }
    
    // implement lhs > rhs in terms of lhs < rhs
    bool Integer::operator>(const Integer& rhs) const {
        return rhs < *this;
    }
    
    // implement lhs <= rhs in terms of !(rhs < lhs)
    bool Integer::operator<=(const Integer& rhs) const {
        return !(rhs < *this);
    }
    
    // implement lhs >= rhs in terms of !(lhs < rhs)
    bool Integer::operator>=(const Integer& rhs) const {
        return !(*this < rhs);
    }
    

    14.5 赋值

    • 示例:
    class Fi {
    public:
        Fi(){}
    };
    
    class Fee {
    public:
        Fee(int) {}
        Fee(const Fi&){}
    };
    
    int main() {
        Fee fee = 1;    // Fee(int)
        Fi fi;
        // 拷贝构造
        Fee fum = fi;   // Fee(Fi)
        // 赋值
        fum = fi;
    }
    

    14.5.1 Assignment Operator

    • Must be a member function
    • Will be generated for you if you don't provide one
      • Same behavior as automatic copy constructor -- memberwise assignment
    • Check for assignment to self
    • Be sure to assign to all data members
    • Return a reference to *this
    T& T::operator=(const T& rhs) {
        // check for self assignment
        if (this != &rhs) {
            // perform assignment
        }
        return *this;
    }
    

    14.6 类型转换

    • A conversion operator can be used to convert an object of one class into
      • an object of another class
      • a built-in type
    • Compilers perform implicit conversions using:
      • Single-argument constructors
      • implicit type conversion operators

    14.6.1 Single-argument constructors

    • 示例:
    class PathName {
        string name;
    public:
        PathName(const string&);
        ~PathName();
    };
    
    int main()
    {
        string abc("abc");
        PathName xyz(abc);  // OK, 拷贝构造
        xyz = abc;          // OK, abc => PathName
    }
    
    
    // 示例二:
    class One {
    public:
        One(){}
    };
    
    class Two {
    public:
        Two(const One&) {}
    };
    
    void f(Two) {}
    
    int main() {
        One one;
        f(one);     // Wants a Two, has a One
    }
    

    14.6.2 Preventing implicit conversions

    • explicit 关键字
    • 示例:
    class PathName {
        string name;
    public:
        explicit PathName(const string&);
        ~ PathName();
    };
    
    int main()
    {
        string abc("abc");
        PathName xyz(abc);  // OK!
        xyz = abc;          // Error!
    }
    

    14.6.3 Conversion operations

    • Operator conversion

      • Function will be called automatically
      • Return type is same as function name
    • 格式:X::operator T()

      • Operator name is any type descriptor
      • No explicit arguments
      • No return type
      • Compiler will use it as a type conversion from X => T
    • 示例:

    class Rational {
    public:
        operator double() const;    // Rational to double
    };
    
    Rational::operator double() const {
        return numrator_/(double)denominator_;
    }
    
    int main()
    {
        Rational r(1, 3);
        double d = 1.3 * r; // r => double
    }
    

    参考资料:

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  • 原文地址:https://www.cnblogs.com/linkworld/p/16385178.html
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