C++類實例聲明無匹配函數

Question

我正在C++中練習OOP，並且正在創建一個銀行帳戶類。下面是類的定義：

#include <iostream> 
#include <cmath> 
#include <limits> 
#include <string> 
#include <cstring> 
#include <sstream> 
#include <stdio.h> 
#include <stdlib.h> 
#include <wchar.h> 

using namespace std; 

const int acc_num_length = 8; 
const int acc_id_num_length = 13; 

class bankAccount { 
private: 
    int acc_num; // account number 
    int acc_id_num; // account holder ID number 
    string acc_tel_no;  // account holder telephone number 
    string acc_first_name; // account holder first name 
    string acc_last_name; // account holder last name 
public: 
    /* constructor */ 
    bankAccount() { 
     acc_num = 0; 
     acc_id_num = 0; 
     acc_tel_no = ""; 
     acc_first_name = ""; 
     acc_last_name = ""; 
    } 
    /* detail retrieval */ 
    int get_acc_num() { 
     return(acc_num); 
    } 
    int get_acc_id_num() { 
     return(acc_id_num); 
    } 
    string get_acc_tel_no() { 
     return(acc_tel_no); 
    } 
    string get_acc_first_name() { 
     return(acc_first_name); 
    } 
    string get_acc_last_name() { 
     return(acc_last_name); 
    } 
}; 

現在在我的主程序我想聲明這個類的一個實例，

int main() { 

    float acc_num = 0; 
    while (((cout << "Enter account number: ") && !(cin >> acc_num)) || (floor(acc_num) < acc_num)) { 
     cout << "Invalid account number.\n"; 
     cin.clear(); 
     cin.ignore(numeric_limits<streamsize>::max(), '\n'); 
    } 
    int acc_num_int = acc_num; 
    string acc_num_string = to_string(acc_num_int); 
    while (acc_num_string.length() != acc_num_length) { 
     cout << "Invalid account number (8 characters only).\n"; 
     cin.clear(); 
     cin.ignore(numeric_limits<streamsize>::max(), '\n'); 
     cin >> acc_num; 
     cout << "Enter account number: \n"; 
    } 

    bankAccount ba1 = bankAccount(acc_num_int,0, "","",""); 
    bankAccount ba2 = bankAccount(); 

    return 0; 
} 

現在一切正常，除了我的實例聲明，

C:/Cartrack/C++/Creating a form/Form/main.cpp: In function 'int main()': C:/Cartrack/C++/Creating a form/Form/main.cpp:68:67: error: no matching function for call to 'bankAccount::bankAccount(int&, long long int, const char [1], const char [1], const char [1])'
bankAccount ba1 = bankAccount(acc_num_int,0, "","",""); ^C:/Cartrack/C++/Creating a form/Form/main.cpp:25:2: note: candidate: bankAccount::bankAccount() bankAccount() { ^~~~~~~~~~~ C:/Cartrack/C++/Creating a form/Form/main.cpp:25:2: note: candidate expects 0 arguments, 5 provided C:/Cartrack/C++/Creating a form/Form/main.cpp:16:7: note: candidate: bankAccount::bankAccount(const bankAccount&) class bankAccount { ^~~~~~~~~~~ C:/Cartrack/C++/Creating a form/Form/main.cpp:16:7: note: candidate expects 1 argument, 5 provided C:/Cartrack/C++/Creating a form/Form/main.cpp:16:7: note: candidate: bankAccount::bankAccount(bankAccount&&) C:/Cartrack/C++/Creating a form/Form/main.cpp:16:7: note: candidate expects 1 argument, 5 provided mingw32-make.exe[1]: * [Form.mk:97: Debug/main.cpp.o] Error 1 mingw32-make.exe[1]: Leaving directory 'C:/Cartrack/C++/Creating a form/Form' mingw32-make.exe: * [Makefile:5: All] Error 2 ====1 errors, 6 warnings====

我不確定爲什麼沒有匹配的電話。我通過了兩個整數和三個字符串，這似乎是大部分人讀過現有威脅時所犯的錯誤。

任何人都可以幫忙嗎？ 謝謝！

Zane

Answer 1

您需要根據需要聲明參數化構造函數。你所擁有的只是默認的（或零參數）構造函數。 請閱讀任何C++書籍/教程的構造函數部分。

而且，你不應該像

bankAccount ba1 = bankAccount(acc_num_int,0, "","",""); 

上面的語句將創建一個無名的對象創建對象並做複製BA1的建設。請閱讀有關複製構造函數。當你需要一個單獨的對象時，這是一個額外函數調用的重載。相反，你可以像下面

bankAccount ba1(acc_num_int,0, "","","");