为什么我不能成功地将文件数据读取到结构数组

时间:2013-04-22 18:29:48

标签: c++ struct deserialization

我正在为我的C ++决赛写一个基于战舰棋盘游戏的程序,我目前正在研究的问题是如何管理用户帐户。我必须允许用户在每次玩游戏时使用相同的帐户并跟踪他们的赢/输记录。我能够将数据写入文件,但是当玩家重新登录然后对其进行排序以查找其用户名时,我需要将其读回。我被困在这一部分。

这是我正在使用的文件,其读作用户名,获胜,损失:

Rocky 0 0
Bob 0 0
dave 0 0
Jerry 0 0
Bert 0 0
Ernie 0 0
Marcus 0 0

编辑:这是我正在重复多次的输出

-858993460
-858993460

-858993460
-858993460

-858993460
-858993460

UserData是一个结构

//begin create/find account function
userData createAccount(userData ud){

    //local variables
    int playerOption;

    //creates object to open files
    ifstream infile;

    //creates object to open files
    ofstream outfile;

    do {
        cout << "Do you have an existing account?" << endl;
        cout << "" << endl;
        cout << "Enter 1 for yes or 2 for no:" << endl;
        cin >> playerOption;
    }
    while (playerOption >= 3 || playerOption <= 0);



    if (playerOption == 1){
        cout << "Enter user name:" << endl;
        cin >> ud.name;

        //opens file in read mode
        infile.open("userData.dat");

        //tests to make sure the file is open
        if (!infile){
            cout << "File open failure!";
        }


        //creates array of user data
        userData userDataArray [SIZE];

        //reads data from file into array until end of file
        int i=0;
        while(i<SIZE){
            infile >> userDataArray[i].name;
            infile >> userDataArray[i].wins;
            infile >> userDataArray[i].losses;
            i++;
        }

        ///test output
        int j=0;
        while (j<SIZE){
            cout << userDataArray[j].name << endl;
            cout << userDataArray[j].wins << endl;
            cout << userDataArray[j].losses << endl;
            j++;
        }
        //end test output


        //closes file
        infile.close();
    }

    else if(playerOption == 2){
        cout << "Enter user name:" << endl;
        cin >> ud.name;
        ud.wins = 0;
        ud.losses = 0;

        //opens file in write mode
        outfile.open("userData.dat",ios::app);

        //tests to make sure the file is open
        if (!outfile){
            cout << "File open failure!";
        }

        //writes userData struct to file
        outfile << ud.name << " " << ud.wins << " " << ud.losses << endl;

        //closes file
        outfile.close();
    }

    return ud;

//end create/find account function
}

1 个答案:

答案 0 :(得分:1)

这是一个最小的例子,它完美地运作:

#include <iostream>
#include <fstream>

using namespace std;

typedef struct { string name; int wins; int losses; } userData;

void createAccount(){
    ifstream infile;
    infile.open("userData.dat");

    userData userDataArray[3];

    for(int i = 0; i < 3; ++i){
        infile >> userDataArray[i].name;
        infile >> userDataArray[i].wins;
        infile >> userDataArray[i].losses;
    }

    for(int i = 0; i < 3; ++i){
        cout << userDataArray[i].name << endl;
        cout << userDataArray[i].wins << endl;
        cout << userDataArray[i].losses << endl;
    }
}

int main(){
    createAccount();
}

输出:

Rocky
0
0
Bob
0
0
dave
0
0

所以你应该逐位简化你的代码,直到它工作。或者从像我这样的简单代码开始,然后逐步实现您所需的功能。

您最初的问题是“为什么我不能成功地将文件数据读取到结构数组”,但显然这不是问题。

首先,SIZE的价值是多少?如果您说输出很长,则可能将size设置为远高于文件中序列化数据的值,并且您正在打印大量未初始化的数据。