输入函数的逻辑错误答案

【问题标题】：Logical Error with input function输入函数的逻辑错误
【发布时间】：2016-02-08 00:53:40
【问题描述】：

我正在做一个关于继承的项目，其中 DollarAmount 是 SpendingRecord 类的父类。这两个输入函数似乎存在问题，因为每当我输入 -1（分隔符）时，它都不会在 main 中输入 if 语句，因此程序永远不会中断并且它接受 -1 作为可接受的美元金额（它如果用户不再希望输入美元金额，则应仅接受 -1 ）。 getDollar、getCent 和 getDELIM 是 DollarAmount 类的函数，所以我假设 SpendingRecord 继承了它们。任何人都可以帮忙吗？

void DollarAmount::inputDollar( istream &ins )
{
        char decimal;
        int dollar, cent;

        cout << "\n$PrintDollar " << endl;
        cout << "Enter the expenditure record (e.g., 1.95, coffee, enter -1 to end): $";

        ins >> dollar;

        if ( dollar != getDELIM())
        {
                ins >> decimal >> cent;
        }
                //
                //check to make sure that the user inputs the correct amount
                //ensures that dollar is in between 0 and 9999
                //ensures that the user does not put 0 dollars and 0 cents
                //ensures that the cents amount is between 0 and 99
                //
        while ( !setCent(cent) || !setDollar(dollar) || dollar == 0 && cent == 0 && dollar != getDELIM())
        {
                cout << "Invalid dollar/cent amount, please reenter: $";
                ins >> dollar >> decimal >> cent;
        }
        while( ins.fail() ){
                ins.clear();
                ins.ignore(1000,'\n');
                cout << "Wrong input types, please reenter: $";
                ins >> dollar >> decimal >> cent;
        }

        setDollar(dollar);
        setCent(cent);
}
void SpendingRecord::inputDollar( istream & in )
{
        string itemName;
        DollarAmount::inputDollar(in);

        if ( dollar != getDELIM() )
        {
                in >> itemName;
                getline( in, itemName );
                SpendingRecord::itemName = itemName;
        }
}

  //in main
      SpendingRecord *ptr = NULL;

        ptr = new SpendingRecord[MAX];

        for ( int i = 0; i < psize; i++ )
        {
                cin >> ptr[i];
                //
                //if user has put in the last element of the array
                //allocate a new array of a larger size and set it to NULL
                //copy the contents of smaller array into larger array
                //delete the old array
                //assign address location of new array to older array
                //increase the physical size of the array
                //
                if ( i == psize - 1 )
                {
                        SpendingRecord *tmp = NULL;
                        tmp = new SpendingRecord[psize*2];
                        for ( int j = 0; j < psize; j++ )
                                tmp[i] = ptr[i];

                        delete [] ptr;
                        ptr = tmp;
                        psize *= 2;
                }
                //if the user enters -1, break and do not include the last element ( i.e. -1 ). The program does not seem to enter this break statement
                if ( ptr[i].getDollar() == getDELIM() && ptr[i].getCent() == 0 )
                {
                        numElements = i;
                        break;
                }
    }

【问题讨论】：

标签： c++ inheritance operator-overloading

【解决方案1】：

什么类型返回函数getDELIM()？应该是int，因为我希望避免溢出。无论如何，它在哪个循环中无限？对我来说，循环中的条件

while (!setCent(cent) || !setDollar(dollar) || dollar == 0 && cent == 0 && dollar != getDELIM())
{ cout << "Invalid dollar/cent amount, please reenter: $"; }

看起来很奇怪，评论对我来说更具可读性。我猜setCent() 在值小于零时可能会返回 false。我认为您在这种情况下缺少括号。

据此我会这样写：

while ((!setCent(cent) || !setDollar(dollar)) || (dollar == 0 && cent == 0) || dollar != getDELIM())
{...}

【讨论】：