更长的获取线cin输入的问题

4 回答

• 0

  一个直接的问题是你从 [1,n] 索引，而C（ std::string ， std::vector ，但也是C样式数组）使用 [0,n) . 这意味着'll access beyond the end of the string. And you'将字符存储到带有 temp[x] 的 temp 中，尽管 temp 的大小始终为0.这两种都是未定义的行为，这可能会产生任何影响（包括崩溃程序） .

  在开发代码时应该使用标准库的调试模式 . 在Visual Studios中，我认为这是默认值;使用g，您需要将 -D_GLIBCXX_CONCEPT_CHECKS -D_GLIBCXX_DEBUG -D_GLIBCXX_DEBUG_PEDANTIC 添加到命令行 .

  处理此问题的最简单方法是复制到新字符串中，随时进行更改：

  std::string results;
  for ( auto current = mening.cbegin(); current != mening.cend(); ++ current ) {
      switch ( *current ) {
      case 'a':
          results += "aoa";
          break;
  
      case 'A':
          results += "AoA";
          break;
  
      default:
          results += *current;
          break;
      }
  }
  

  如果你确实想要进行替换，那就太棘手了 . 当您插入的文本多于开头时，迭代器将失效 . 所以你需要这样的东西：

  static std::string const Ao( "Ao" );
  static std::string const ao( "ao" );
  for ( auto current = mening.begin(); current != mening.end(); ++ current ) {
      switch ( *current ) {
      case 'a':
          current = mening.insert( current, ao.begin(), ao.end() ) + 2;
          break;
  
      case 'A':
          current = mening.insert( current, Ao.begin(), Ao.end() ) + 2;
          break;
      }
  }
  

  就个人而言，我赞成复制到一个新的字符串 .

  回复于 2024-05-02T10:57:31+08:00
• 0

  我建议你应该使用标准函数 std::string::insert 来插入字符 'oA' 或 'oa' . 它将使您的代码更易于处理和调试 .

  或者你可以这样做：

  #include <iostream>
  #include <string>
  
  using namespace std;
  
  int main(int argc, const char * argv[])
  {
  string mening, temp; //The mening string is the word/sentence the user will input.
  int play = 1, add;
  
  while (play == 1) {
  cout<<"Type in the sentence: ";
  
  getline(cin, mening); //The input is saved in the string variable mening.
  
  unsigned long y = mening.size(); //Grabs the amount of characters in input; this number is saved in the unsigned long variable y.
  add = 0; //Makes sure the int variable add is reset to 0 if the loop restarts.
  
  for (int n = 0; n<y; n++ ) {
      cout << mening[n];
      if (mening[n] == 'a' || mening[n] == 'A') 
          cout << "o" << mening[n];
  }
  
  cout<<endl<<"Do you want to do it again? (yes/no): ";
  
  getline(cin, mening);
  
      cin.clear();
      cout << flush;
      cout.flush();
      cout.clear();
  
  if (mening == "Yes" || mening == "yes" || mening == "YES") {
  
  }
  else {
      play = 2;
  }
  }
  
  
  cout<<endl<<"The program will now close.";
  
  return 0;
  }
  

  回复于 2024-05-02T10:57:31+08:00
• 0

  #include <iostream>
  #include <string>
  
  
  std::string ReplaceA(std::string s) {
      std::string temp = "";
      for (unsigned int k = 0; k < s.size(); k++) {
          if (s[k] == 'a' || s[k] == 'A') {
              temp = temp + s[k];
              temp = temp + "o";
              temp = temp + s[k];
              }
              else { 
                  temp = temp + s[k];
              }
  
          }
      return temp;
  }
  
  int main(int argc, const char * argv[])
  {
      std::string mening; //The mening string is the word/sentence the user will input.
      int play = 1;
  
      while (play == 1) {
          std::cout<<"Type in the sentence: ";
  
          getline(std::cin, mening); //The input is saved in the string variable mening.
  
          std::cout << ReplaceA(mening) << std::endl;
  
          std::cout << "Do you want to do it again? (yes/no): ";
  
          std::cin >> mening;
          if( std::cin.fail() || ( mening != "yes" && mening != "no" ) ) {
              std::cout << "Bad Input\nDo you want to do it again? (yes/no): ";
              std::cin.clear();
              std::cin.ignore('256','\n');
              std::cin >> mening;
          }else{
              if(mening == "no") break;
          }
          std::cin.clear();
          std::cin.ignore('256','\n');
  
      }
  
  
      std::cout << "The program will now close.";
  
      return 0;
  }
  

  更有条理，具有处理返回字符串的转换的函数 .

  你也可以尝试这个功能，两个都工作，一个似乎有点漂亮 .

  std::string ReplaceA(std::string s) {
      std::string temp = "";
      for(std::string::iterator k = s.begin(); k != s.end(); k++) {
          switch(*k) {
          case 'a': temp += "aoa"; break;
          case 'A': temp += "AoA"; break;
          default: temp += *k; break;
          }
      }
      return temp;
  }
  

  回复于 2024-05-02T10:57:31+08:00
• 1

  实际上有一个尾随字符'\ n'来自最后一个cin >>，所以getline取这个'\ n'字符并被终止;在getline之前忽略缓冲区的内容......

  cin.ignore（）;
  getling（cin，string_name）

  回复于 2024-05-02T10:57:31+08:00