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;
}

我有一个复杂的数字程序，当我想显示复数时，它显示错误，因为（4-1i）应该显示为（4-i）。虽然（4-1i）在技术上是正确的，但它不会显示为我希望它显示。我在print方法中创建了一个简单的if语句，但它不起作用，因为变量imaginary不是一个字符串。 *如何使我的变量imaginary等于1，它返回＆＃34;＆＃34;或空白或没有任何东西，以便它可以打印出适当的＆＃34; i）＆＃34;我已经设定了。

示例输出：

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)

注意差异是（-2 + 1i）......我不喜欢这样。我不想要那个。另外，我不想要那个（0-32i）。所以基本上当它是0或者它是1时我希望打印功能能够反映出来。所以差异看起来像（-2 + i），而第二复杂Num Squared看起来像（32i）

现在我的代码：

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
};

和司机：

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;
}

带输出

(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; }

}

指定-1i，1i，0i，0real，0i和0real 最终的if-else块。除此之外别无他法。

Answer 3

首先，当你检查两件事是否相等时，你想要使用==或.equals（）。在这种情况下，您需要==，因为您要比较基元接下来，此函数是void函数。这意味着您已将其定义为不返回任何内容，但您尝试返回空字符串。
您想要做的是在imaginary为1时打印一个模式，在imaginary为1时打印不同的模式1.使用打印语句，就像在默认情况下一样，打印imaginary = 1的特殊情况。

如何回归＆＃34;＆＃34;当float的值为1时，OR为空

3 个答案: