通过将结构体传递给函数来添加复数的 C++ 程序

该程序将两个复数作为结构体，并使用函数将它们相加。

要理解此示例，您应该具备以下 C++ 编程主题的知识：

示例：将两个复数相加的源代码

// Complex numbers are entered by the user

#include <iostream>
using namespace std;

typedef struct complex {
    float real;
    float imag;
} complexNumber;

complexNumber addComplexNumbers(complex, complex);

int main() {
    complexNumber num1, num2, complexSum;
    char signOfImag;

    cout << "For 1st complex number," << endl;
    cout << "Enter real and imaginary parts respectively:" << endl;
    cin >> num1.real >> num1.imag;

    cout << endl
         << "For 2nd complex number," << endl;
    cout << "Enter real and imaginary parts respectively:" << endl;
    cin >> num2.real >> num2.imag;

    // Call add function and store result in complexSum
    complexSum = addComplexNumbers(num1, num2);

    // Use Ternary Operator to check the sign of the imaginary number
    signOfImag = (complexSum.imag > 0) ? '+' : '-';

    // Use Ternary Operator to adjust the sign of the imaginary number
    complexSum.imag = (complexSum.imag > 0) ? complexSum.imag : -complexSum.imag;

    cout << "Sum = " << complexSum.real << signOfImag << complexSum.imag << "i";

    return 0;
}

complexNumber addComplexNumbers(complex num1, complex num2) {
    complex temp;
    temp.real = num1.real + num2.real;
    temp.imag = num1.imag + num2.imag;
    return (temp);
}

输出

For 1st complex number,
Enter real and imaginary parts respectively:
3.4
5.5

For 2nd complex number,
Enter real and imaginary parts respectively:
-4.5
-9.5
Sum = -1.1-4i

在这个程序中，用户输入的两个复数存储在结构体 num1 和 num2 中.

这两个结构体被传递给 addComplexNumbers() 计算总和并将结果返回给 main() 函数。

这个结果存储在结构体 complexSum 中.

然后，确定和的虚部的符号存储在 char 变量 signOfImag 中.

// Use Ternary Operator to check the sign of the imaginary number
signOfImag = (complexSum.imag > 0) ? '+' : '-';

如果 complexSum 的虚部是正的，那么 signOfImag 被赋值为 '+' 。否则，它被赋值为 '-' 。

然后我们调整值 complexSum.imag.

/// Use Ternary Operator to adjust the sign of the imaginary number
complexSum.imag = (complexSum.imag > 0) ? complexSum.imag : -complexSum.imag;

此代码更改 complexSum.imag，如果发现它是负值，则为正。

这是因为如果它是负的，那么将它与 signOfImag 将在输出中给我们两个负号。

因此，我们将值更改为正以避免符号重复。

在此之后，我们最终显示总和。