c++ - 重载后递增运算符的复数平方值，无需运算符实例化

Question

这是输出:

First Complex Number:
Enter real part of complex number: 3
Enter imaginary part of complex number: 6

Second Complex Number:
Enter real part of complex number: 5
Enter imaginary part of complex number: -5

a == (-27.00+36.00i)
b == (5.00-5.00i)
a+b == (-22.00+31.00i)
a-b == (-32.00+41.00i)
a*b == (45.00+315.00i)
a*a == (-567.00-1944.00i)
b*b == (0.00-50.00i)
a*a (using postincrement) ==(-27.00+36.00i)

如您所见，并非所有涉及a 的内容都是错误的，因为它将a(复数)的平方作为a .因此，虽然“a*a (using postincrement) == (-27.00+36.00i)”的答案是正确的答案......它说“a == ( -27.00+36.00i)"是不正确的，因为它应该是 a==(3+6i)。我相信错误在于我的代码的重载和友元方面，但是我不确定如何修复它，因为我没有收到任何错误...这是我代码中的逻辑问题。

这是我的代码:

#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 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'
        << "a*a (using postincrement) ==" << a++ << '\n';
        cout << endl;

    system("PAUSE");
}

Answer 1

这不是关于运算符重载，而是关于 order of evaluation .

在声明中:

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'
        << "a*a (using postincrement) ==" << a++ << '\n';

参数将按照您的预期从左到右打印，但计算每个参数的各个步骤不必按该顺序完成。

编译器可以按任何顺序计算每个子表达式的值，只要该值在需要打印时“准备好”即可。所以它可以做这样的事情:

auto&& temp1 = a + b;
auto&& temp2 = a - b;
auto&& temp3 = a*b;
auto&& temp4 = a*a;
auto&& temp5 = b*b;
auto&& temp6 = a++;
cout << fixed << setprecision(2)
        << "a == " << a << '\n'
        << "b == " << b << '\n'
        << "a+b == " << temp1 << '\n'
        << "a-b == " << temp2 << '\n'
        << "a*b == " << temp3 << '\n'
        << "a*a == " << temp4 << '\n'
        << "b*b == " << temp5 << '\n'
        << "a*a (using postincrement) ==" << temp6 << '\n';

如果发生这种情况，您将获得预期的行为，因为子表达式将按从左到右的顺序求值。但是当前的 C++ 标准 (C++14) 允许编译器重新排序评估，如果它需要更少的堆栈空间和更少的寄存器，这可能允许它更好地优化代码。该语句的另一个有效执行是:

auto&& temp1 = a++;
auto&& temp2 = b*b;
auto&& temp3 = a*a;
auto&& temp4 = a*b;
auto&& temp5 = a - b;
auto&& temp6 = a + b;
cout << fixed << setprecision(2)
        << "a == " << a << '\n'
        << "b == " << b << '\n'
        << "a+b == " << temp6 << '\n'
        << "a-b == " << temp5 << '\n'
        << "a*b == " << temp4 << '\n'
        << "a*a == " << temp3 << '\n'
        << "b*b == " << temp2 << '\n'
        << "a*a (using postincrement) ==" << temp1 << '\n';

这次你可以看到a++操作先发生，所以后面的计算将在发生后进行。

编译器还可以选择子表达式求值的任何其他顺序，例如它可以评估 a*a 然后 ++a 然后 a-b 等(但在实践中最常见的顺序是从左到右和从右到左)

---

顺便说一句，您的代码还有另外两个严重的问题:

此函数返回对局部变量的引用:

ComplexNum& ComplexNum::getComplexNum()
{
    ComplexNum keyboard;
    // ...
    return keyboard; 
}

该局部变量在函数返回后不再存在，因此任何使用该引用的尝试都会访问一个不存在的对象。 永远不要这样做!您的编译器应该警告您此处存在问题，启用编译器的警告并且不要忽略它们!

其次，真的重载像++这样的运算符来表示完全不同的东西，比如将一个对象乘以自身，这不是一个好主意。只有当运算符对您要执行的操作有意义时，才应使用运算符重载。递增不是平方，所以这是一个糟糕的选择。我会为此定义一个普通的(非运算符)函数。