访问结构时出现分段错误

Question

程序一直运行，直到它检查用户输入的名称。当您输入您希望在从充满客户信息的文件中导入的结构数组中搜索的名称时，它会返回分段错误核心转储。这让我很困惑。

#include <iostream>
#include <string>
#include <fstream>
#include <cstring>
using namespace std;

struct AccountsDataBase{

        char name[50];
        string email;
        long int phone;
        string address;
};


#define MAX 80

AccountsDataBase * account = new AccountsDataBase[MAX];


void readIn(ifstream& file){
        int i=0;
        while(!file.eof()){
                file >> account[i].name >> account[i].email >> account[i].phone >> account[i].address;
        }
}

void getAccount(){

        char userPick[50];
        char streamName[50];

        cout << " What account will we  be using? " << endl;

        cin.getline(streamName, 50);

        for(int i=0; strcmp(account[i].name, streamName)!=0; i++){
                if( strcmp(account[i].name, streamName)==0){
                        cout << "\n\n FOUND IT!! \n\n";
                        cout << account[i].name << "\n" << account[i].email << "\n" << account[i].phone << "\n" << account[i].address << endl;
                }
        }
}

int main(){
        ifstream file;
        file.open("2.dat"); //opens data account records text
        readIn(file);
        getAccount();
        delete account;
        return 0;
}

Answer 1

您的循环不断将所有内容读入数组的初始元素：

while(!file.eof()){
    file >> account[i].name >> account[i].email >> account[i].phone >> account[i].address;
}  

因为i 的值永远不会增加。您可以将其转换为 for 循环，如下所示：

for (count = 0 ; count < MAX && !file.eof() ; count++) {
    file >> account[count].name >> account[count].email >> account[count].phone >> account[count].address;
}

请注意，我将i 更改为count：

AccountsDataBase * account = new AccountsDataBase[MAX];
int count = 0;

这将帮助您解决另一个问题 - 确定数组何时以 getAccount 函数结束。目前，您假设记录始终存在，因此外部循环继续进行。现在你有了count，你可以像这样改变循环：

for(int i=0; i < count && strcmp(account[i].name, streamName)!=0; i++){
    if( strcmp(account[i].name, streamName)==0){
        cout << "\n\n FOUND IT!! \n\n";
        cout << account[i].name << "\n" << account[i].email << "\n" << account[i].phone << "\n" << account[i].address << endl;
        break;
    }
}
if (i == count) {
    cout << "Not found." << endl;
}