需要幫助擦除並在C++中插入

Question

所以基本上我想創建一個接受字符串的格式函數，並將該字符串中的單詞替換爲用戶想要替換的任何字符。起初，我有一些問題與不可判定的迭代器，直到我意識到當你改變一個字符串的大小時，你可以使任何迭代器失效。現在它不會再拋出異常現在輸出與輸入相同。任何建議？

string& formatFn(string& s, string& oldWord, string& newWord) 
{ 
    string word = ""; 
    for (auto iter1 = s.begin(); iter1 != s.end(); ++iter1) 
    {  
      string tmpWord = "";       
     if (!isblank(*iter1))  // Testing for whitespace 
     { 
      tmpWord += *iter1; 
      if (tmpWord == oldWord) 
      { 
       string::iterator beg = iter1 - word.size(); 
       string::iterator end = iter1; 
       auto sIter = s.erase(beg, end);     // Get the position returned by erase 
       auto i = sIter - s.begin();      // Get an index 
       s = s.insert(s[i], newWord); 
      } 
     } 
     if (isblank(*iter1)) 
     { 
      tmpWord.clear(); 
     } 
    } 
    return s; 
} 

Answer 1

string::iterator beg = iter1 - word.size(); 

我不知道什麼word實際上做。您正試圖刪除oldWord，對不對？然後，它應該是：

string::iterator beg = iter1 - oldWord.size(); 

編輯：這是你的代碼的改進版本：

string formatFn(const string& s, const string& oldWord, const string& newWord) { 
    string result = "";  // holds the string we want to return 
    string word = "";  // while iterating over 's', holds the current word 

    for (auto iter1 = s.begin(); iter1 != s.end(); ++iter1) {  
     if (!isblank(*iter1)) 
      word += *iter1; 
     else { // if it is a whitespace, it must be the end of some word 
      // if 'word' is not same as 'oldword', just append it 
      // otherwise append 'newWord' instead 

      if (word == oldWord) 
       result += newWord; 
      else 
       result += word; 
      result += *iter1; 
      word.clear(); // reset 'word' to hold the next word in s 
     } 
    } 

    // the end of the string might not end with a whitespace, so the last word 
    // might be skipped if you don't make this test 

    if (word == oldWord) 
     result += newWord; 
    else 
     result += word; 

    return result; 
} 

Answer 2

如果您已經使用字符串爲什麼不`噸使用的所有方法？

for (auto it = text.find(o_text); it != string::npos; it = text.find(o_text)){ 
    text.replace(it, o_text.size(), n_text); 
} 

Answer 3

您是過於複雜的：

std::string replace_all(std::string s, const std::string& sOld, const std::string& sNew) 
{ 
    std::size_t p = s.find(sOld); 
    while (p != std::string::npos) 
    { 
     s.replace(p, sOld.length(), sNew); 
     p = s.find(sOld, p + sNew.length()); 
    } 
    return s; 
} 

如果您正在尋找只能更換整個單詞（當前您的圖謀是不會做）：

#include <iostream> 
#include <string> 

bool test(const std::string& s, const std::string& sOld, std::size_t pos) 
{ 
    return (pos == 0 || !::isalpha(s[pos - 1])) && (!::isalpha(s[pos + sOld.length()]) || pos + sOld.length() >= s.length()); 
} 

std::size_t find_word(const std::string& s, const std::string& sOld, std::size_t pos) 
{ 
    pos = s.find(sOld, pos); 
    while (pos != std::string::npos && (!test(s, sOld, pos) && pos < s.length())) 
    { 
     pos++; 
     pos = s.find(sOld, pos); 
    } 
    return pos; 
} 

std::string replace_all(std::string s, const std::string& sOld, const std::string& sNew) 
{ 
    std::size_t p = find_word(s, sOld, 0); 
    while (p != std::string::npos && p < s.length()) 
    { 
     s.replace(p, sOld.length(), sNew); 
     p = find_word(s, sOld, p + sNew.length()); 
    } 
    return s; 
} 

int main() 
{ 
    std::string sOrig = "eat Heat eat beat sweat cheat eat"; 
    std::string sOld = "eat"; 
    std::string sNew = "ate"; 
    std::string sResult = replace_all(sOrig, sOld, sNew); 
    std::cout << "Result: " << sResult << std::endl; 
    // Output: "ate Heat ate beat sweat cheat ate" 
    return 0; 
}