模板化运算符创建一个错误，指出该运算符不明确

Question

我已经定义了一个模板化的运算符来满足运算符参数中的两种类型的输入。 getDataFromStream() 出现错误，如何定义运算符以消除这种歧义？

BankAccount.h

template <typename T>
istream& operator>>( istream&, T&);     //input operator

template <typename T>
istream& operator>>( istream& is, T& aBankAccount) {
//get BankAccount details from stream
    return ( aBankAccount.getDataFromStream( is));
}

BankAccount.cpp

包括“BankAccount.h”

void BankAccount::readInBankAccountFromFile( const string& fileName) {
    ifstream fromFile;
    fromFile.open( fileName.c_str(), ios::in);  //open file in read mode
    if ( fromFile.fail())
        cout << "\nAN ERROR HAS OCCURED WHEN OPENING THE FILE.";
    else
        fromFile >> (*this);    //read  all info from bank account file
                 ^^
    fromFile.close();           //close file: optional here
}

但也在这个提供错误的函数中 (BankAccount.cpp)

istream& BankAccount::getDataFromStream( istream& is) {
//get BankAccount details from stream
    is >> accountType_;                     //get account type
    is >> accountNumber_;                   //get account number
    is >> sortCode_;                        //get sort code
    is >> creationDate_;                    //get creation date
    is >> balance_;                         //get balance_
    is >> transactions_;                    //get all transactions (if any)
    return is;
}

Cashpoint.cpp #includes "Cashpoint.h"（包含#includes "BankAccount.h"）

bool CashPoint::linkedCard( string cashCardFileName) const {
//check that card is linked with account data
    ifstream inFile;
    inFile.open( cashCardFileName.c_str(), ios::in);    //open file
    bool linked(false);
    if ( ! inFile.fail()) //file should exist at this point 
    {   //check that it contain some info in addition to card number
        string temp;
        inFile >> temp; //read card number
        inFile >> temp; //ready first account data or eof
        if (inFile.eof())
            linked = false;
        else
            linked = true;
        inFile.close();         //close file: optional here
    }
    return linked;
}

编辑：操作员的 cashpoint.cpp 和 bankaccount.cpp 中的错误>> 可能与将 BankAccount.h 包含在 Cashpoint.h 中有关

Answer 1

您重载operator>> 的方法对我来说似乎很奇怪。

template <typename T>
istream& operator>>( istream& is, T& aBankAccount) {
//get BankAccount details from stream
    return ( aBankAccount.getDataFromStream( is));
}

你基本上是在说：不管T是什么类型，调用一个方法getDataFromStream就可以了。这没有任何意义！它还会产生模棱两可的重载，因为当T 为std::string 时，编译器现在有两个operator>> 可能调用：一个来自您的代码，一个来自标准库。

您似乎只想为 BankAccount 类型添加重载。

istream& operator>>(istream& is, BankAccount& aBankAccount) {
    return ( aBankAccount.getDataFromStream( is));
}