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
}
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。