# C++ 程序：通過將結構傳遞給函數來相加復數 > 原文： [https://www.programiz.com/cpp-programming/examples/complex-number-add](https://www.programiz.com/cpp-programming/examples/complex-number-add) #### 該程序將兩個復數作為結構并通過使用函數將它們相加。 要理解此示例，您應該了解以下 [C++ 編程](/cpp-programming "C++ tutorial")主題： * [C++ 結構](/cpp-programming/structure) * [C++ 結構和函數](/cpp-programming/structure-function) * * * * * * ## 示例：相加兩個復數的源代碼 ```cpp // 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 n1, n2, temporaryNumber; char signOfImag; cout << "For 1st complex number," << endl; cout << "Enter real and imaginary parts respectively:" << endl; cin >> n1.real >> n1.imag; cout << endl << "For 2nd complex number," << endl; cout << "Enter real and imaginary parts respectively:" << endl; cin >> n2.real >> n2.imag; signOfImag = (temporaryNumber.imag > 0) ? '+' : '-'; temporaryNumber.imag = (temporaryNumber.imag > 0) ? temporaryNumber.imag : -temporaryNumber.imag; temporaryNumber = addComplexNumbers(n1, n2); cout << "Sum = " << temporaryNumber.real << temporaryNumber.imag << "i"; return 0; } complexNumber addComplexNumbers(complex n1,complex n2) { complex temp; temp.real = n1.real+n2.real; temp.imag = n1.imag+n2.imag; return(temp); } ``` **輸出** ```cpp 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 ``` 在問題中，用戶輸入的兩個復數存儲在結構`n1`和`n2`中。 這兩個結構傳遞給`addComplexNumbers()`函數，該函數計算總和并將結果返回給`main()`函數。 最后，通過`main()`函數顯示總和。