c++ - 当我将数组声明为 3 时，为什么我的数组只接受 2 个值？

Question

这是我的代码:

    int amount;
    cout<<"How many books are you donating? ";
    cin>>amount;
    string donation[amount];

    cout<<"Enter the titles:"<<endl;
    for (int i=0; i<amount; i+=1)
    {
        getline(cin.ignore(numeric_limits<streamsize>::max(), '\n'), donation[i]);
    }

    cout<<endl<<"You donated these:";
    for (int j=0; j<amount; j+=1)
    {
        cout<<donation[j]<<endl;
    }

即使我只为 amount 输入 3 个值，这也接受 5 个值.然后它只打印值 1、3 和 5。

如果我把它写成 getline(cin.ignore(), donation[i]);或 getline(cin.ignore(1, '\n'), donation[i]); ，那么输出变成这样:

How many books are you donating? 3
Enter the titles:
book1
book2
book3

You donated these:
book1
ook2
ook3

我应该在 cin.ignore()中写什么让它忽略 \n值(value)？

Answer 1

您需要在第一个 cin>>amount 读取后丢弃\n，

注意:你可以使用 i++ 或++i 而不是 i+=1 (这是一种更标准的方式)

#include <iostream>
#include <string>
#include <limits>

using namespace std;

int main()
{
    int amount;
    cout<<"How many books are you donating? ";
    cin>>amount;
    string donation[amount];
    cin.ignore (std::numeric_limits<std::streamsize>::max(), '\n'); 

    cout<<"Enter the titles:"<<endl;
    for (int i=0; i<amount; i++)
    {
        cout<< "Enter title "<<i << ":";
        getline(cin, donation[i]);
    }

    cout<<endl<<"You donated these:"<<endl;
    for (int j=0; j<amount; j++)
    {
        cout<<donation[j]<<endl;
    }

    return 0;
}