虽然它们是相同的(对我来说),但是在工作方面被击中的类功能是错过的

时间:2016-02-06 21:33:37

标签: c++

我试图运行这个类函数字符串,每个函数都在类头中声明并在class.cpp中定义

我遇到的问题是,它会跳过街道名称,然后将其放入城市(将所有内容移到一个城市),然后当涉及到邮政编码时,只需重新输入街道号码即可。

class.h看起来像

class AddressBook
{
    private:
    string firstName;
    string lastName;
    int streetNum;
    string streetName;
    string city;
    string state;
    int zipCode;

    public:
    static int entryCnt;

    void setFirstName(string temp);
    void setLastName(string temp);
    void setStreetNum(int tempInt);
    void setStreetName(string temp);
    void setCity(string temp);
    void setState(string temp);
    void setZipCode(int tempInt);

    //copies some properties into out agruments
    void getFirstName(string buff, int sz) const;
    void getLastName(string buff, int sz) const;
    void addEntryFromConsole();
    void printToConsole(int entryCnt);
    void appendToFile();

    void operator=(const AddressBook& obj);

};

bool operator==(const AddressBook& obj1, const AddressBook& obj2);

#endif // !ADDRESSBOOK_ENTRY

string temp;
    int tempInt;

和相关的class.cpp部分看起来像

#include <iostream>
#include "AddressBook.h"


void AddressBook::setFirstName(string temp) {
    firstName = temp;
}


void AddressBook::setLastName(string temp) {
    lastName = temp;
}

void AddressBook::setStreetNum(int tempInt) {
    streetNum = tempInt;
}

void AddressBook::setStreetName(string temp) {
    streetName = temp;
}
void AddressBook::setCity(string temp) {
    city = temp;
}
void AddressBook::setState(string temp) {
    state = temp;   
}
void AddressBook::setZipCode(int tempInt) {
    zipCode = tempInt;
}

和我的main.cpp部分。

while (openFile.good())
{
    getline(openFile, temp);
    AddrBook[entryCnt].setFirstName(temp);
    openFile.clear();

    getline(openFile, temp);
    AddrBook[entryCnt].setLastName(temp);
    openFile.clear();

    openFile >> tempInt;
    //getline(openFile, tempInt);
    AddrBook[entryCnt].setStreetNum(tempInt);
    openFile.clear();

    getline(openFile, temp);
    AddrBook[entryCnt].setStreetName(temp);
    openFile.clear();

    getline(openFile, temp);
    AddrBook[entryCnt].setCity(temp);
    openFile.clear();

    getline(openFile, temp);
    AddrBook[entryCnt].setState(temp);
    openFile.clear();

    openFile >> tempInt;
    AddrBook[entryCnt].setZipCode(tempInt);
    openFile.clear();

    entryCnt = entryCnt + 1;
}

提前感谢您的帮助!

1 个答案:

答案 0 :(得分:0)

使用

导致命中和遗漏问题
openFile >> tempInt;

接着是

getline(openFile, temp);

读取tempInt后,换行符仍留在流中。下一次调用getline只会获取一个空字符串。

您可以在致电阅读getline后立即向tempInt添加另一个电话。

openFile >> tempInt;
std::string ignoreString;
getline(openFile, ignoreString);

...

getline(openFile, temp);
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