C++ 程序通过将结构传递给函数来添加复数-立地货

通过将结构传递给函数来添加复数的 C++ 程序

该程序以两个复数为结构，使用函数相加。

要理解此示例，您应该了解以下C++ 编程主题:

C++ 结构

C++ 结构和函数

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

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 num1和 num2中。

这两个结构被传递给 addComplexNumbers() 函数，该函数计算总和并将结果返回给 main() 函数。

这个结果存储在结构 complexSum中。

然后，确定和的虚部的符号并将其存储在 char变量 signOfImag中。

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

如果 complexSum 的虚部为正，则 signOfImag 被赋值为 '+'。否则，它被分配值 '-'。

然后我们调整 complexSum.imag的值。

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

如果发现 complexSum.imag 为负值，此代码会将其更改为正值。

这是因为如果它是负数，那么将它与 signOfImag 一起打印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，我们终于显示了总和。