重載的預遞增運算符錯誤“多個運算符++匹配了這些操作數”

Question

我在代碼中重載了postincrement（a ++）運算符，並且嘗試在代碼中重載preincrement運算符（++ a）以執行對復數“ a”進行平方的函數，但是，只有postincrement運算符在此時此刻。 我應該明確指出，前后遞增運算符都調用相同的函數-它們都應使用平方方法/函數。 因此，盡管他們擔任過職位，但他們應該給我同樣的答案。

調用時，preincrement運算符下面有一條紅線：“有多個運算符++匹配這些操作數。但是我明確地考慮了“ ++（）”和“ ++（int）”，以解決這兩個問題++ a和a ++。

我想念什么嗎？ 這是我的代碼。

#include<iostream>
#include<iomanip>
using namespace std;

class ComplexNum
{
public:
    ComplexNum(float = 0.0, float = 0.0); //default constructor that uses default arg. in case no init. are in main
    ComplexNum& getComplexNum(); //get real and imaginary numbers from keyboard
    ComplexNum& sum(ComplexNum a, ComplexNum b); //method to add two ComplexNum numbers together
    ComplexNum& diff(ComplexNum a, ComplexNum b); //method to find the difference of two complex numbers
    ComplexNum& prod(ComplexNum a, ComplexNum b); //method to find the product of two complex numbers
    ComplexNum& square(ComplexNum a); //method to find square using pre/post increment operators

    //overloaded operators
    ComplexNum& operator =  (const ComplexNum& that) = default;
    ComplexNum& operator += (const ComplexNum& that) { return sum(*this, that); }
    ComplexNum& operator -= (const ComplexNum& that) { return diff(*this, that); }
    ComplexNum& operator *= (const ComplexNum& that) { return prod(*this, that); }
    ComplexNum& operator ++() { return square(*this); } //called for ++num
    ComplexNum operator ++(int) { return square(*this); } //called for num++

    ostream& print(ostream& stm = cout) const;


private:
    float real; //float data member for real number (to be entered in by user)
    float imaginary; //float data member for imaginary number (to be entered in by user)

    //non-member overloaded operators
    //a is passed by value
    friend ComplexNum operator+ (ComplexNum a, const ComplexNum& b) { return a += b; }
    friend ComplexNum operator- (ComplexNum a, const ComplexNum& b) { return a -= b; }
    friend ComplexNum operator* (ComplexNum a, const ComplexNum& b) { return a *= b; }
    friend ComplexNum operator++(ComplexNum a) { return a++; } //friend for num ++
    friend ComplexNum operator++(ComplexNum a) { return ++a; } //friend for ++num

    friend ostream& operator<< (ostream& stm, const ComplexNum& c) { return c.print(stm); }
};

ComplexNum::ComplexNum(float a, float b)
{
    real = a;
    imaginary = b;
}

ComplexNum& ComplexNum::getComplexNum()
{
    ComplexNum keyboard;
    cout << "Enter real part of complex number: ";
    cin >> real;

    cout << "Enter imaginary part of complex number: ";
    cin >> imaginary;

    return keyboard; 
}

ComplexNum& ComplexNum::square(ComplexNum a)
{
    this->real = (a.real * a.real) - (a.imaginary * a.imaginary);
    this->imaginary = (2 * (a.real * a.imaginary));
    return *this;
}

ComplexNum& ComplexNum::sum(ComplexNum a, ComplexNum b)
{
    this->real = a.real + b.real;
    this->imaginary = a.imaginary + b.imaginary;
    return *this;
}

ComplexNum& ComplexNum::diff(ComplexNum a, ComplexNum b)
{
    this->real = a.real - b.real;
    this->imaginary = a.imaginary - b.imaginary;
    return *this;
}

ComplexNum& ComplexNum::prod(ComplexNum a, ComplexNum b)
{
    this->real = (a.real * b.real) - (a.imaginary * b.imaginary);
    this->imaginary = (a.real * b.imaginary) + (b.real * a.imaginary);
    return *this;
}

ostream& ComplexNum::print(ostream& stm) const
{
    return stm << "(" << noshowpos << real << showpos << imaginary << "i)";
}

int main()
{
    ComplexNum a, b;
    cout << "First Complex Number:" << endl;
    a.getComplexNum();
    cout << endl;
    cout << "Second Complex Number:" << endl;
    b.getComplexNum();
    cout << endl;
    cout << fixed << setprecision(2)
        << "a == " << a << '\n'
        << "b == " << b << '\n'
        << "a+b == " << a + b << '\n'
        << "a-b == " << a - b << '\n'
        << "a*b == " << a*b << '\n'
        << "a*a == " << a*a << '\n'
        << "b*b == " << b*b << '\n';
        cout << "a*a (using postincrement) == " << a++ << '\n'; //works fine
        cout << "a*a (using preincrement) == " << ++a << '\n'; //gives me a [more than one operator error] underneath "++" 
        cout << endl;

    system("PAUSE");
}

Answer 1

非成員預遞增簽名為

T& operator++(T& a); // ++a

非成員后增量為

T operator++(T& a, int); // a++

您必須將這些定義更改為

friend ComplexNum& operator++(ComplexNum& a, int) { return a++; } //friend for num ++
friend ComplexNum operator++(ComplexNum& a) { return ++a; } //friend for ++num