动态数组大小并在 getline() 处崩溃；

Question

我最近一直在开发一个程序，该程序将名称作为输入并最终对它们进行排序和二进制搜索。但是，在尝试使数组成为动态大小（每次循环迭代都会增加一）时，它遇到了各种问题。

我可以制作由 20 个元素组成的字符串数组，并且程序可以运行，但我的任务的额外功劳是使其具有动态大小。目前，程序一旦到达“getline(cin, Names[x]);”就会崩溃，没有任何类型的错误代码。 我一直在四处寻找，我知道在这种情况下使用向量而不是数组会更容易，但是我不相信我可以在这个分配中使用向量。

谢谢

原始代码

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

void main()
{
    int x = 0;
    string * Names = new string[x];
    bool NameInputEnd(0);

    cout << "    Enter your names to be sorted\n";
    cout << "To exit just press [Enter] at any time\n";

    do
    {
        cout << x << endl;
        cout << "\n< Name " << (x + 1) << " > = ";

        !!**CRASHES HERE**!!

        getline(cin, Names[x]);

        if (Names[x].empty() || x == 19)
        {
            cout << "\nFinal Name Amount = " << (x + 1) << endl << endl;
            NameInputEnd = 1;
            continue;
        }

        x++;

    } while (NameInputEnd == 0);

    delete [] Names;
}

变化

int tempsize(1), x(0);
string * Names = new string[tempsize];
...

do
{
...
    x++;
    tempsize++;
}while (NameInputEnd == 0);

Answer 1

数组一旦创建就不能调整大小。您必须销毁它并使用现有数据的副本创建一个新数组。例如：

#include <iostream>
#include <string>
#include <algorithm>

using namespace std;

void main()
{
    int x = 0;
    int capacity = 20;
    string * Names = new string[capacity];
    string Name;

    cout << "    Enter your names to be sorted\n";
    cout << "To exit just press [Enter] at any time\n";

    do
    {
        cout << x << endl;
        cout << "\n< Name " << (x + 1) << " > = ";

        if ((!getline(cin, Name)) || Name.empty())
            break;

        if (x == capacity)
        {
            int newCapacity = capacity + 20;
            string *newNames = new string[newCapacity];
            copy(Names, Names + x, newNames);
            delete [] Names;
            Names = newNames;
            capacity = newCapacity;
        }

        Names[x] = Name;
        ++x;    
    }
    while (true);

    cout << "\nFinal Name Amount = " << x << endl << endl;

    delete [] Names;
}

不过，您确实应该使用std::vector：

#include <iostream>
#include <string>
#include <vector>

using namespace std;

void main()
{
    vector<string> Names;
    string Name;

    Names.reserve(20); // optional

    cout << "    Enter your names to be sorted\n";
    cout << "To exit just press [Enter] at any time\n";

    do
    {
        cout << Names.size() << endl;
        cout << "\n< Name " << (Names.size() + 1) << " > = ";

        if ((!getline(cin, Name)) || Name.empty())
            break;

        Names.push_back(Name);
    }
    while (true);

    cout << "\nFinal Name Amount = " << Names.size() << endl << endl;
}