c++ - C++ 程序的双输入验证问题

Question

所以我试图：

1. 在 do while 循环中进行双重输入验证
2. 检查是否已经进行了移动以及它是否是逻辑输入。(#1-9)

我最初认为 if else 语句，但我不确定如何使用 else 语句回到循环的开头。

do
{
  cout << "Interesting move, What is your next choice?: ";
  cin >> play;
  Pused[1] = play;
  if(play != Pused[0] && play != cantuse[0] && play != cantuse[1] )
  {
    switch(play)
    {
      default:cout << "Your choice is incorrect\n\n";
      break;
    }   
  }
  else
  { }
}    
while(play != 1 && play != 2 && play != 3 && play != 4 && play != 5 && play != 6 && play != 7 && play != 8 && play != 9);
Dis_board(board);

Answer 1

使用“继续”关键字返回到循环的开始。

Answer 2

只需删除else. 我认为这不是必需的。如果满足条件，您的循环将自动继续while。

Answer 3

您的问题有点难以理解，但是在此循环中您需要解决一些条件：

1. 询问用户输入
2. 检查用户输入是否有效（1-9之间且之前未使用过）
3. 如果我们有一个有效的选择，则退出循环

所以我们需要记录已经完成的动作并检查用户的输入是否在有效的选择范围内，我们可以使用一个只有在选择了有效的选择时才退出的循环。

int choice;
bool used[9] = { false }; // Set all values to false
std::cout << "Interesting move, what is your next choice?: ";
do {
    std::cin >> choice;
    // Here we check if the choice is within the range we care about
    // and not used, note if the first condition isn't true then
    // the second condition won't be evaluated, so if choice is 11
    // we won't check used[10] because the && will short circuit, this
    // lets us avoid array out of bounds. We also need to
    // offset the choice by 1 for the array because arrays in C++
    // are indexed from 0 so used runs from used[0] to used[8]
    if((choice >= 1 && choice <= 9) && !used[choice - 1]) {
        // If we're here we have a valid choice, mark it as used
        // and leave the loop
        used[choice - 1] = true;
        break; // Exit the loop regardless of the while condition
    }

    // If we've made it here it means the above if failed and we have
    // an invalid choice. Restart the loop!
    std::cout << "\nInvalid choice! Input: ";
} while (true);