函数采用填充结构的数组,并返回具有相同结构的更大数组

时间:2013-07-11 20:26:47

标签: c++

#include <iostream>
#include <string>

using namespace std;

struct FriendInfo
    {
        string name;
        int last_day;
    };

FriendInfo GrowArray(FriendInfo friend_list,int sizing);

int main()
{
    int initial_friends;
    cout << "How many friends would you like to add--> ";
    cin >> initial_friends;
    FriendInfo friends[initial_friends];
    for (int i = 0; i < initial_friends;i++)
    {
        cout << "What is your friend's name?--> ";
        cin >> friends[i].name;
        cout << "When was the last time you talked to " << friends[i].name << " ?--> ";
        cin >> friends[i].last_day;
    }
    for (int i = 0; i < initial_friends;i++)
        cout << friends[i].name << " " << friends[i].last_day << "\n";
    cout << "What would you like to do now?\n1. Add another friend?\n2. Update one of yourfriends?\n3. Sort friends by day?4. Quit.\n--> ";
    int option;
    cin >> option;
    while (option != 4)
    {
        if (option == 1)
        {
            friends = GrowArray(friends,initial_friends);
            cout << "What is your new friend's name?--> ";
            cin >> friends[initial_friends].name;
            cout << "When was the last time you talked to " << friends[initial_friends].name << " ?--> ";
            cin >> friends[initial_friends].last_day;
        }

    }
}

FriendInfo GrowArray(FriendInfo friend_list, int sizing)
{
     FriendInfo new_list[sizing + 1];
     for (int i = 0;i < sizing;i++)
     {
         new_list[i].name = friend_list[i].name;
         new_list[i].last_day = friend_list.last_day;
 }
 return new_list;
}

这个程序将结构放入一个包含朋友姓名的数组,以及他们与他们交谈的最后一天。稍后的选项之一是添加另一个朋友。函数GrowArray使用朋友和日期获取初始数组,使另一个数组生成额外的点并将原始数组复制到新数组中。但是,当我使用该功能时,我收到此错误 - &gt;错误:无法将'(FriendInfo *)(&amp; friends)'从'FriendInfo *'转换为'FriendInfo'。怎么了?

1 个答案:

答案 0 :(得分:1)

你不能像那样重新分配friends,因为它不是指针,它是一个数组。 即使你可以,你所做的事情也不安全,因为GrowArray只是在堆栈上创建一个新的数组,当函数返回时它将被销毁。

您应该使用newdelete来创建和销毁数组(然后您可以将它们作为指针传递),或者最好使用处理所有这些东西的std::vector对你而言。