Question

在这段代码中减少我的条件的最好方法是什么，是简单地实现Boost库以进行字符串比较还是有另一种方法呢？如果可能的话，我会更希望避免使用Boost。

如何在以下代码中实现所说的问题？

cout << " Please Enter An Answer Of Yes(Y) or No(N):";
string question; cin >> question;

if (question == "yes" || question == "Yes" || question == "Y" || question == "y" || question == "YEs" || question == "YES" || question == "yeS" || question == "yES") 
{
    cout << "You said yes!" << endl;
    return 1;
} 
else if (question == "No" || question == "NO" || question == "nO" || question == "N" || question == "n")
{
    cout << "You said no!" <<endl;
    return 0;
}  
else 
{
    AskAQuestion();
}

Answer 1

这可能不是最有效的解决方案，但是如果将整个字符串转换为小写，则只需要检查单词和字母，而不需要检查所有可能的排列。所以如果你有

void make_lowercase(std::string& data)
{
    std::transform(data.begin(), data.end(), data.begin(), ::tolower);
}

然后

cout << " Please Enter An Answer Of Yes(Y) or No(N):";
string question; cin >> question;

if (question == "yes" || question == "Yes" || question == "Y" || question == "y" || question == "YEs" || question == "YES" || question == "yeS" || question == "yES") 
//...

成为

cout << " Please Enter An Answer Of Yes(Y) or No(N):";
string question; 
cin >> question;
make_lowercase(question);

if (question == "yes" || question == "y")
//...

Answer 2

以下内容并不完全相同，因为您没有在第一个if语句中检查“ yEs”（可能的情况），但这可能与您要查找的内容大致相同。它是基于较低的用户在回答答案之前先回答用户的答案，以减少要检查的案例数。

#include <algorithm>
#include <cctype>

...

std::cout << "Please Enter An Answer Of Yes(Y) or No(N): ";
std::string answer; std::cin >> answer;
std::transform(answer.begin(), answer.end(), answer.begin(), std::tolower);

if (answer == "yes" || answer == "y")
{
  std::cout << "You said yes!\n";
  return 1;
}
else if (answer == "no" || answer == "n")
{
  std::cout << "You said no!\n";
  return 0;
}
else
{
  AskAQuestion();
}

我自由地将question重命名为answer，在这种情况下似乎更合乎逻辑。

条件语句简化-C ++

2 个答案: