C++ &引用；“边做边做”；循环挂起，不重复

我遇到了一个do while循环不重复的问题。 当我在提示再次添加两个数字时选择N时，它会按预期退出程序，但选择Y或Y会挂起程序，将我带到下一行，但使我无法键入任何内容。如果你能帮助我，我将不胜感激。谢谢

#include <iostream>
using namespace std;
int main()
{

int number1, number2; //Two numbers
int sum;              //Sum
char again;     //For yet another attempt


{  //Inputting the two numbers

    cout << "\nEnter two numbers, and i will add them: ";
    cin >> number1 >> number2;

    //Adding the two numbers

    sum = number1 + number2;
    cout << "The sum of the two numbers you have picked is " << sum << "\n\n";

    //Does the user want to add two other numbers?
    cout << "Do you want to add two other numbers? (Y/N) ";
    cin >> again;
}
    while (again == ('y') || again == ('Y'));


   return 0;
}

#包括
使用名称空间std；
int main（）
{
int number1，number2；//两个数字
int sum；//sum
再次char；//再次尝试
{//输入两个数字
cout>number1>>number2；
//把这两个数字相加
总和=数字1+数字2；
库特
当循环语法在上面时，我猜你们在语法上错过了do世界。
试着替换你所做的，而块如下

do{  //Inputting the two numbers

cout << "\nEnter two numbers, and i will add them: ";
cin >> number1 >> number2;

//Adding the two numbers

sum = number1 + number2;
cout << "The sum of the two numbers you have picked is " << sum << "\n\n";

//Does the user want to add two other numbers?
cout << "Do you want to add two other numbers? (Y/N) ";
cin >> again;}while (again == ('y') || again == ('Y'));

do{//输入两个数字
cout>number1>>number2；
//把这两个数字相加
总和=数字1+数字2；
是的，你需要一段时间做一件事。说得再好不过了

每次在运行程序之前向代码或调试中添加新内容时，也要进行生成检查，并检查所有警告或错误（如果有）。此外，还要依靠用于帮助查找小错误的IDE。我们有时都会忘记一些小错误。
谢谢！我真的很惊讶我错过了这些！对不起，我是新手。这是“回答你的问题”按钮吗？对不起，我没有理解你若要接受答案，是否按“回答您的问题”按钮？若要将答案标记为已接受，请单击答案旁边的复选标记，将其从灰色变为已填写。代码是否应阻塞（用于循环）在
while
关键字之后，而不是之前？这是一个正常的while语句。我只是忘记了do部分。编码时需要遵循语法
do{  //Inputting the two numbers

cout << "\nEnter two numbers, and i will add them: ";
cin >> number1 >> number2;

//Adding the two numbers

sum = number1 + number2;
cout << "The sum of the two numbers you have picked is " << sum << "\n\n";

//Does the user want to add two other numbers?
cout << "Do you want to add two other numbers? (Y/N) ";
cin >> again;}while (again == ('y') || again == ('Y'));