用另一个子字符串 C++ 替换子字符串

Posted

技术标签:

【中文标题】用另一个子字符串 C++ 替换子字符串【英文标题】:Replace substring with another substring C++ 【发布时间】:2011-06-06 07:45:19 【问题描述】:

如何用 C++ 中的另一个子字符串替换字符串中的子字符串,我可以使用哪些函数?

eg: string test = "abc def abc def";
test.replace("abc", "hij").replace("def", "klm"); //replace occurrence of abc and def with other substring

【问题讨论】:

几乎是 ***.com/questions/3418231/… 的副本,在接受的答案中有更强大的解决方案。 【参考方案1】:

C++ 中没有一个内置函数可以做到这一点。如果您想用另一个子字符串替换一个子字符串的所有实例,您可以通过混合调用string::findstring::replace 来实现。例如:

size_t index = 0;
while (true) 
     /* Locate the substring to replace. */
     index = str.find("abc", index);
     if (index == std::string::npos) break;

     /* Make the replacement. */
     str.replace(index, 3, "def");

     /* Advance index forward so the next iteration doesn't pick it up as well. */
     index += 3;

在这段代码的最后一行,我将index 增加了插入字符串的长度。在这个特定的示例中 - 用 "def" 替换 "abc" - 这实际上不是必需的。但是,在更一般的设置中,跳过刚刚被替换的字符串很重要。例如,如果您想用"abcabc" 替换"abc",而不跳过新替换的字符串段,此代码将不断替换部分新替换的字符串,直到内存耗尽。独立地,无论如何跳过这些新字符可能会稍微快一些,因为这样做可以通过 string::find 函数节省一些时间和精力。

希望这会有所帮助!

【讨论】:

我认为您不需要增加索引,因为您已经替换了数据,因此无论如何它都不会拾取它。 @Aidiakapi 如果将其转换为通用函数,它不会陷入无限循环,因为它会将搜索位置 (index) 推进到被替换的字符串部分. @TimR。你说得对,我在回复 rossb83,他说索引的增加是不必要的。只是想防止错误信息。所以对于其他所有人:将索引增加替换字符串的长度(在本例中为 3)是必要的。不要从代码示例中删除它。 @rossb83 cmets 需要清理或说明。有 5 个赞成的评论说您不必增加索引,然后一个人说您需要以粗体表示。这对来这里学习的人没有帮助。 @JulianCienfuegos 我刚刚更新了解决这个问题的答案 - 感谢您指出这一点! (另外,Aidiakapi 是其他人……不确定那是谁。)【参考方案2】:

如果您确定字符串中存在所需的子字符串,那么这会将第一次出现的"abc" 替换为"hij"

test.replace( test.find("abc"), 3, "hij");

如果你在测试中没有“abc”,它会崩溃,所以小心使用它。

【讨论】:

【参考方案3】:
using std::string;

string string_replace( string src, string const& target, string const& repl)

    // handle error situations/trivial cases

    if (target.length() == 0) 
        // searching for a match to the empty string will result in 
        //  an infinite loop
        //  it might make sense to throw an exception for this case
        return src;
    

    if (src.length() == 0) 
        return src;  // nothing to match against
    

    size_t idx = 0;

    for (;;) 
        idx = src.find( target, idx);
        if (idx == string::npos)  break;

        src.replace( idx, target.length(), repl);
        idx += repl.length();
    

    return src;

由于它不是 string 类的成员,因此它不允许像您的示例中那样漂亮的语法,但以下将做等效:

test = string_replace( string_replace( test, "abc", "hij"), "def", "klm")

【讨论】:

【参考方案4】:
    string & replace(string & subj, string old, string neu)
    
        size_t uiui = subj.find(old);
        if (uiui != string::npos)
        
           subj.erase(uiui, old.size());
           subj.insert(uiui, neu);
        
        return subj;
    

我认为这符合您的要求,只需很少的代码!

【讨论】:

您没有考虑多次出现/替换【参考方案5】:

Boost String Algorithms Library方式:

#include <boost/algorithm/string/replace.hpp>

 // 1. 
  string test = "abc def abc def";
  boost::replace_all(test, "abc", "hij");
  boost::replace_all(test, "def", "klm");



 // 2.
  string test = boost::replace_all_copy
  (  boost::replace_all_copy<string>("abc def abc def", "abc", "hij")
  ,  "def"
  ,  "klm"
  );

【讨论】:

杰伊。我需要 boost 来替换所有子字符串。 Boost 大多是矫枉过正。【参考方案6】:

替换子字符串应该没有那么难。

std::string ReplaceString(std::string subject, const std::string& search,
                          const std::string& replace) 
    size_t pos = 0;
    while((pos = subject.find(search, pos)) != std::string::npos) 
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    
    return subject;

如果你需要性能,这里有一个优化的函数,它修改输入字符串,它不会创建字符串的副本:

void ReplaceStringInPlace(std::string& subject, const std::string& search,
                          const std::string& replace) 
    size_t pos = 0;
    while((pos = subject.find(search, pos)) != std::string::npos) 
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    

测试:

std::string input = "abc abc def";
std::cout << "Input string: " << input << std::endl;

std::cout << "ReplaceString() return value: " 
          << ReplaceString(input, "bc", "!!") << std::endl;
std::cout << "ReplaceString() input string not changed: " 
          << input << std::endl;

ReplaceStringInPlace(input, "bc", "??");
std::cout << "ReplaceStringInPlace() input string modified: " 
          << input << std::endl;

输出:

Input string: abc abc def
ReplaceString() return value: a!! a!! def
ReplaceString() input string not modified: abc abc def
ReplaceStringInPlace() input string modified: a?? a?? def

【讨论】:

需要添加检查 if (search.empty()) return; 以避免在传递空“搜索”时出现无限循环。 尝试了 ReplaceString 功能 - 无效。但回答如下: str.replace(str.find(str2),str2.length(),str3);很简单,效果很好。【参考方案7】:

如果替换字符串的长度与要替换的字符串的长度不同,我认为所有解决方案都会失败。 (搜索“abc”并替换为“xxxxxx”) 一般的方法可能是:

void replaceAll( string &s, const string &search, const string &replace ) 
    for( size_t pos = 0; ; pos += replace.length() ) 
        // Locate the substring to replace
        pos = s.find( search, pos );
        if( pos == string::npos ) break;
        // Replace by erasing and inserting
        s.erase( pos, search.length() );
        s.insert( pos, replace );
    

【讨论】:

【参考方案8】:

概括 rotmax 的答案,这里是搜索和替换字符串中所有实例的完整解决方案。如果两个子字符串的大小不同,则使用 string::erase 和 string::insert. 替换子字符串,否则使用更快的 string::replace。

void FindReplace(string& line, string& oldString, string& newString) 
  const size_t oldSize = oldString.length();

  // do nothing if line is shorter than the string to find
  if( oldSize > line.length() ) return;

  const size_t newSize = newString.length();
  for( size_t pos = 0; ; pos += newSize ) 
    // Locate the substring to replace
    pos = line.find( oldString, pos );
    if( pos == string::npos ) return;
    if( oldSize == newSize ) 
      // if they're same size, use std::string::replace
      line.replace( pos, oldSize, newString );
     else 
      // if not same size, replace by erasing and inserting
      line.erase( pos, oldSize );
      line.insert( pos, newString );
    
  

【讨论】:

谢谢。使用几年没有问题;然而,最终需要将oldStringnewString 都变成const 参数。【参考方案9】:
str.replace(str.find(str2),str2.length(),str3);

在哪里

str 是基本字符串 str2 是要查找的子字符串 str3 是替换子串

【讨论】:

这只会替换第一次出现,不是吗? 我建议确保 str.find(str2) 的结果不等于 std::string::npos auto found = str.find(str2); if(found != std::string::npos) str.replace(found, str2.length(), str3); 我不打算用这个来编写整个应用程序,但是在没有对输入进行任何检查的情况下,有些情况是未定义的......【参考方案10】:

@Czarek Tomczak 改进的版本。 允许std::stringstd::wstring

template <typename charType>
void ReplaceSubstring(std::basic_string<charType>& subject,
    const std::basic_string<charType>& search,
    const std::basic_string<charType>& replace)

    if (search.empty())  return; 
    typename std::basic_string<charType>::size_type pos = 0;
    while((pos = subject.find(search, pos)) != std::basic_string<charType>::npos) 
         subject.replace(pos, search.length(), replace);
         pos += replace.length();
    

【讨论】:

【参考方案11】:

在c++11,你可以使用std::regex_replace

#include <string>
#include <regex>

std::string test = "abc def abc def";
test = std::regex_replace(test, std::regex("def"), "klm"); // replace 'def' -> 'klm'
// test = "abc klm abc klm"

【讨论】:

如果我们有 c++11 那就太好了!! 小心这不能很好地概括,你最终可能会传递一些 std:regex 解释不同的东西,比如 std::regex_replace(test, std::regex("."), "klm") ...【参考方案12】:
std::string replace(const std::string & in
                  , const std::string & from
                  , const std::string & to)
  if(from.size() == 0 ) return in;
  std::string out = "";
  std::string tmp = "";
  for(int i = 0, ii = -1; i < in.size(); ++i) 
    // change ii
    if     ( ii <  0 &&  from[0] == in[i] )  
      ii  = 0;
      tmp = from[0]; 
     else if( ii >= 0 && ii < from.size()-1 )  
      ii ++ ;
      tmp = tmp + in[i];
      if(from[ii] == in[i]) 
       else 
        out = out + tmp;
        tmp = "";
        ii = -1;
      
     else 
      out = out + in[i];
    
    if( tmp == from ) 
      out = out + to;
      tmp = "";
      ii = -1;
    
  
  return out;
;

【讨论】:

【参考方案13】:

这是一个使用递归的解决方案,它用另一个子字符串替换所有出现的子字符串。无论字符串的大小如何,这都有效。

std::string ReplaceString(const std::string source_string, const std::string old_substring, const std::string new_substring)

    // Can't replace nothing.
    if (old_substring.empty())
        return source_string;

    // Find the first occurrence of the substring we want to replace.
    size_t substring_position = source_string.find(old_substring);

    // If not found, there is nothing to replace.
    if (substring_position == std::string::npos)
        return source_string;

    // Return the part of the source string until the first occurance of the old substring + the new replacement substring + the result of the same function on the remainder.
    return source_string.substr(0,substring_position) + new_substring + ReplaceString(source_string.substr(substring_position + old_substring.length(),source_string.length() - (substring_position + old_substring.length())), old_substring, new_substring);

使用示例:

std::string my_cpp_string = "This string is unmodified. You heard me right, it's unmodified.";
std::cout << "The original C++ string is:\n" << my_cpp_string << std::endl;
my_cpp_string = ReplaceString(my_cpp_string, "unmodified", "modified");
std::cout << "The final C++ string is:\n" << my_cpp_string << std::endl;

【讨论】:

【参考方案14】:

这是我使用构建器策略编写的解决方案:

#include <string>
#include <sstream>

using std::string;
using std::stringstream;

string stringReplace (const string& source,
                      const string& toReplace,
                      const string& replaceWith)

  size_t pos = 0;
  size_t cursor = 0;
  int repLen = toReplace.length();
  stringstream builder;

  do
  
    pos = source.find(toReplace, cursor);

    if (string::npos != pos)
    
        //copy up to the match, then append the replacement
        builder << source.substr(cursor, pos - cursor);
        builder << replaceWith;

        // skip past the match 
        cursor = pos + repLen;
    
   
  while (string::npos != pos);

  //copy the remainder
  builder << source.substr(cursor);

  return (builder.str());

测试:

void addTestResult (const string&& testId, bool pass)

  ...


void testStringReplace()

    string source = "123456789012345678901234567890";
    string toReplace = "567";
    string replaceWith = "abcd";
    string result = stringReplace (source, toReplace, replaceWith);
    string expected = "1234abcd8901234abcd8901234abcd890";

    bool pass = (0 == result.compare(expected));
    addTestResult("567", pass);


    source = "123456789012345678901234567890";
    toReplace = "123";
    replaceWith = "-";
    result = stringReplace(source, toReplace, replaceWith);
    expected = "-4567890-4567890-4567890";

    pass = (0 == result.compare(expected));
    addTestResult("start", pass);


    source = "123456789012345678901234567890";
    toReplace = "0";
    replaceWith = "";
    result = stringReplace(source, toReplace, replaceWith);
    expected = "123456789123456789123456789"; 

    pass = (0 == result.compare(expected));
    addTestResult("end", pass);


    source = "123123456789012345678901234567890";
    toReplace = "123";
    replaceWith = "-";
    result = stringReplace(source, toReplace, replaceWith);
    expected = "--4567890-4567890-4567890";

    pass = (0 == result.compare(expected));
    addTestResult("concat", pass);


    source = "1232323323123456789012345678901234567890";
    toReplace = "323";
    replaceWith = "-";
    result = stringReplace(source, toReplace, replaceWith);
    expected = "12-23-123456789012345678901234567890";

    pass = (0 == result.compare(expected));
    addTestResult("interleaved", pass);



    source = "1232323323123456789012345678901234567890";
    toReplace = "===";
    replaceWith = "-";
    result = utils_stringReplace(source, toReplace, replaceWith);
    expected = source;

    pass = (0 == result.compare(expected));
    addTestResult("no match", pass);


【讨论】:

【参考方案15】:
std::string replace(std::string str, std::string substr1, std::string substr2)

    for (size_t index = str.find(substr1, 0); index != std::string::npos && substr1.length(); index = str.find(substr1, index + substr2.length() ) )
        str.replace(index, substr1.length(), substr2);
    return str;

不需要任何额外库的简短解决方案。

【讨论】:

这个问题还有 14 个其他答案。为什么不解释一下为什么你的更好? 看起来是迄今为止最优雅的答案,没有任何过度设计【参考方案16】:
std::string replace(std::string str, const std::string& sub1, const std::string& sub2)

    if (sub1.empty())
        return str;

    std::size_t pos;
    while ((pos = str.find(sub1)) != std::string::npos)
        str.replace(pos, sub1.size(), sub2);

    return str;

【讨论】:

【参考方案17】:
#include <string>

第一:

void replace_first(std::string& text, const std::string& from,
   const std::string& to)

    const auto at = text.find(from, 0);

    if (at != std::string::npos)
        text.replace(at, from.length(), to);

全部:

void replace_all(std::string& text, const std::string& from,
   const std::string& to)

    for (auto at = text.find(from, 0); at != std::string::npos;
        at = text.find(from, at + to.length()))
    
        text.replace(at, from.length(), to);
    

计数:

size_t replace_count(std::string& text,
   const std::string& from, const std::string& to)

    size_t count = 0;

    for (auto at = text.find(from, 0); at != std::string::npos;
        at = text.find(from, at + to.length()))
    
        ++count;
        text.replace(at, from.length(), to);
    

    return count;

复制:

std::string replace_all_copy(const std::string& text,
   const std::string& from, const std::string& to)

    auto copy = text;
    replace_all(copy, from, to);
    return copy;

【讨论】:

以上是关于用另一个子字符串 C++ 替换子字符串的主要内容,如果未能解决你的问题,请参考以下文章

C++编程,查找字符串子串并替换。

数组篇在python中如何查找最长字符串子串

用另一个 NSAttributedString 替换 NSAttributedString 的子字符串

用另一条记录的结果替换空/空字符串

python Python方法用另一种模式替换模式,在字符串中找到

Ruby - 用另一个字符串替换第一次出现的子字符串