C++类实例声明无匹配函数

Question

我正在C++中练习OOP，并且正在创建一个银行帐户类。下面是类的定义：

#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); 
    } 
}; 

现在在我的主程序我想声明这个类的一个实例，

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; 
} 

现在一切正常，除了我的实例声明，

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====

我不确定为什么没有匹配的电话。我通过了两个整数和三个字符串，这似乎是大部分人读过现有威胁时所犯的错误。

任何人都可以帮忙吗？ 谢谢！

Zane

Answer 1

您需要根据需要声明参数化构造函数。你所拥有的只是默认的（或零参数）构造函数。 请阅读任何C++书籍/教程的构造函数部分。

而且，你不应该像

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

上面的语句将创建一个无名的对象创建对象并做复制BA1的建设。请阅读有关复制构造函数。当你需要一个单独的对象时，这是一个额外函数调用的重载。相反，你可以像下面

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