当 float 值为 1 时如何返回“”或为空

Question

void ComplexNum::printComplexNum()
{
    if (imaginary = 1)
    {
        return ""; //cannot return "" I've also tried imaginary == ""; to no avail
    }
    cout << "(" << noshowpos << real << showpos << imaginary << "i)" << endl;
}

我有一个复数程序，当我想显示复数时，它会错误地显示它们，因为 (4-1i) 应该显示为 (4-i)。虽然 (4-1i) 在技术上是正确的，但它并没有像我希望的那样显示。我在 print 方法中创建了一个简单的 if 语句，但它不起作用，因为 imaginary 变量不是字符串。 *如何使我的变量 imaginary 等于 1 时返回 "" 或空白或什么都没有，以便它可以打印出我设置的适当的 "i)"。

样本输出：

First Complex Number:
Enter real part of complex number: 2
Enter imaginary part of complex number: -3
Form '(a+bi)': (2-3i)

Second Complex Number:
Enter real part of complex number: 4
Enter imaginary part of complex number: -4
Form '(a+bi)': (4-4i)

The addition of the two Complex Numbers is: (6-7i)
The difference of the two Complex Numbers is: (-2+1i)
The product of the two Complex Numbers is: (-4-20i)
First Complex Number Squared: (-5-12i)
Second Complex Number Squared: (0-32i) 

注意区别是 (-2+1i)...我不喜欢那样。我不想要那个。另外，我不想要那个（0-32i）。所以基本上当它为 0 或为 1 时，我希望 print 函数能够反映这一点。所以差异看起来像 (-2+i) 而第二个复数平方看起来像 (32i)

现在进入我的代码：

class ComplexNum
{
public:
    ComplexNum(float = 0.0, float = 0.0); //default constructor that uses default arg. in case no init. are in main
    void getComplexNum(); //get real and imaginary numbers from keyboard
    void sum(ComplexNum a, ComplexNum b); //method to add two ComplexNum numbers together
    void diff(ComplexNum a, ComplexNum b); //method to find the difference of two complex numbers
    void prod(ComplexNum a, ComplexNum b); //method to find the product of two complex numbers
    void square(); //squares values of a and b when called in main
    void printComplexNum(); //print sum, diff, prod, square 
    void formComplexNum(); //and "a+bi" form


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)
    float realSquare; //squared real number data member for square method
    float imaginarySquare; //squared imaginary number data member for square method
};

还有司机：

int main()
{
    ComplexNum a, b, c, d, e, f, g;
    cout << "First Complex Number:" << endl;
    a.getComplexNum();
    a.formComplexNum();
    cout << endl;

    cout << "Second Complex Number:" << endl;
    b.getComplexNum();
    b.formComplexNum();
    cout << endl;

    c.sum(a, b);  
    c.printComplexNum();

    d.diff(a, b);
    d.printComplexNum();

    e.prod(a, b);
    e.printComplexNum();

    cout << "First Complex Number Squared: ";
    a.square();
    cout << "Second Complex Number Squared: ";
    b.square();
    cout << endl;

    system("PAUSE");

    return 0;
}

Answer 1

您已将此标记为 C++

(4-1i) 应显示为 (4-i)

您可能会发现 std::stringstream 很有帮助。它简化了对虚部的特殊处理：

virtual int foo()
{
   std::cout << std::endl;
   show(4, -2);
   show(5, -1);

   return(0);
}

void show(int real, int imaginary)
{
   std::stringstream ss; // default is blank
   if      (-1 == imaginary)   { ss << "-i)"; }
   else /* (-1 != imaginary)*/ { ss << imaginary << "i)"; }

   std::cout << "("
             << std::noshowpos << real
             << std::showpos   << ss.str()
             << std::endl;
}

有输出

(4-2i) 
(5-i)

Answer 2

void ComplexNum::printComplexNum()
{
    if (imaginary == -1 && real == 0)
    { cout << "(-i)" << endl; }
    else if (imaginary == 1 && real == 0)
    { cout << "(i)" << endl; }
    else if (imaginary == -1)
    { cout << "(" << noshowpos << real << "-i)" << endl; }
    else if (imaginary == 1)
    { cout << "(" << noshowpos << real << "+i)" << endl; }
    else if (real == 0 && imaginary == 0)
    { cout << "(0)" << endl; }
    else if (real == 0)
    { cout << "(" << noshowpos << imaginary << "i)" << endl; }
    else if (imaginary == 0)
    { cout << "(" << noshowpos << real << ")" << endl; }
    else
    { cout << "(" << noshowpos << real << showpos << imaginary << "i)" << endl; }

}

-1i、1i、0i、0real、0i 和 0real 的帐户 最终的 if-else 块。没有其他方法可以解决这个问题。

Answer 3

首先，当您检查两个事物是否相等时，您希望使用 == 或 .equals()。在这种情况下，您需要 ==，因为您正在比较原语。
接下来，这个函数是一个void函数。这意味着您已将其定义为不返回任何内容，但您正试图返回一个空字符串。
您要做的是在 imaginary 为 1 时打印一个模式，并在 imaginary 不是 1 时打印一个不同的模式。使用 print 语句，就像您在默认情况下所做的那样，打印 @987654323 的特殊情况@ = 1。