How to replace all occurrences of a character in string

Question

What is the effective way to replace all occurrences of a character with another character in std  string

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here s a solution i rolled  in a maximal DRI spirit  it will search sNeedle in sHaystack and replace it by sReplace   nTimes if non 0  else all the sNeedle occurences  it will not search again in the replaced text   std  string str replace      std  string sHaystack  std  string sNeedle  std  string sReplace       size t nTimes 0        size t found   0  pos   0  c   0      size t len   sNeedle size        size t replen   sReplace size        std  string input sHaystack        do           found   input find sNeedle  pos           if  found    std  string  npos                break                    input replace found  len  sReplace           pos   found   replen            c        while  nTimes    c  lt  nTimes        return input

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What about Abseil StrReplaceAll   From the header file      This file defines  absl  StrReplaceAll     a general-purpose string    replacement function designed for large  arbitrary text substitutions     especially on strings which you are receiving from some other system for    further processing  e g  processing regular expressions  escaping HTML    entities  etc     StrReplaceAll  is designed to be efficient even when only    one substitution is being performed  or when substitution is rare        If the string being modified is known at compile-time  and the substitutions    vary   absl  Substitute    may be a better choice        Example        std  string html escaped   absl  StrReplaceAll user input                                                         amp      amp amp                                                          lt      amp lt                                                          gt      amp gt                                                                amp quot                                                               amp  39

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Old School  -   std  string str    H  recursos audio youtube libre falta      for  int i   0  i  lt  str size    i          if  str i                    str i                   std  cout  lt  lt  str    Result      H  recursos audio youtube libre falta

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For simple situations this works pretty well without using any other library then std  string  which is already in use    Replace all occurences of character a with character b in some string   for  size t i   0  i  lt  some string size      i        if  some string i      a             some string replace i  1   b              If the string is large or multiple calls to replace is an issue  you can apply the technique mentioned in this answer  https   stackoverflow com a 29752943 3622300

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std  string doesn t contain such function but you could use stand-alone replace function from algorithm header    include  lt algorithm gt   include  lt string gt   void some func       std  string s    example string     std  replace  s begin    s end     x    y       replace all  x  to  y

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As Kirill suggested  either use the replace method or iterate along the string replacing each char independently    Alternatively you can use the find method or find first of depending on what you need to do  None of these solutions will do the job in one go  but with a few extra lines of code you ought to make them work for you   -

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This works  I used something similar to this for a bookstore app  where the inventory was stored in a CSV  like a  dat file   But in the case of a single char  meaning the replacer is only a single char  e g      it must be in double quotes     in order not to throw an invalid conversion const char     include  lt iostream gt   include  lt string gt   using namespace std   int main         int count   0      for the number of occurences         final hold variable of corrected word up to the npos j     string holdWord              a temp var in order to replace 0 to new npos     string holdTemp              a csv for a an entry in a book store     string holdLetter    Big Java 7th Ed Horstman 978-1118431115 99 85           j   npos     for  int j   0  j  lt  holdLetter length    j               if  holdLetter j                         if   count    0                                             holdWord   holdLetter replace j  1                                          else                    string holdTemp1   holdLetter replace j  1                              since replacement is three positions in length                     must replace new replacement s 0 to npos-3  with                    the 0 to npos - 3 of the old replacement                  holdTemp   holdTemp1 replace 0  j-3  holdWord  0  j-3                     holdWord                        holdWord   holdTemp                             holdTemp                   count                         cout  lt  lt  holdWord  lt  lt  endl      return 0        result  Big Java 7th Ed   Horstman   978-1118431115   99 85   Uncustomarily I am using CentOS currently  so my compiler version is below   The C   version  g     C  98 default   g    GCC  4 8 5 20150623  Red Hat 4 8 5-4  Copyright  C  2015 Free Software Foundation  Inc  This is free software  see the source for copying conditions   There is NO warranty  not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE

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The question is centered on character replacement  but  as I found this page very useful  especially Konrad s remark   I d like to share this more generalized implementation  which allows to deal with substrings as well  std  string ReplaceAll std  string str  const std  string amp  from  const std  string amp  to        size t start pos   0      while  start pos   str find from  start pos      std  string  npos            str replace start pos  from length    to           start pos    to length       Handles case where  to  is a substring of  from            return str     Usage  std  cout  lt  lt  ReplaceAll string  quot Number Of Beans quot    std  string  quot   quot    std  string  quot   quot     lt  lt  std  endl  std  cout  lt  lt  ReplaceAll string  quot ghghjghugtghty quot    std  string  quot gh quot    std  string  quot X quot     lt  lt  std  endl  std  cout  lt  lt  ReplaceAll string  quot ghghjghugtghty quot    std  string  quot gh quot    std  string  quot h quot     lt  lt  std  endl   Outputs   Number Of Beans XXjXugtXty hhjhugthty   EDIT  The above can be implemented in a more suitable way  in case performances are of your concern  by returning nothing  void  and performing the changes directly on the string str given as argument  passed by address instead of by value  This would avoid useless and costly copy of the original string  while returning the result  Your call  then    Code   static inline void ReplaceAll2 std  string  amp str  const std  string amp  from  const std  string amp  to           Same inner code           No return statement     Hope this will be helpful for some others

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For completeness  here s how to do it with std  regex   include  lt regex gt   include  lt string gt   int main         const std  string s    quot example string quot       const std  string r   std  regex replace s  std  regex  quot x quot     quot y quot

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include  lt iostream gt   include  lt string gt  using namespace std     Replace function   string replace string word  string target  string replacement       int len  loop 0      string nword     let      len word length        len--      while loop lt  len           let word substr loop  1           if let  target               nword nword replacement           else              nword nword let                    loop              return nword       Main   int main       string word    cout lt  lt  Enter Word       cin gt  gt word    cout lt  lt replace word   x    y   lt  lt endl    return 0

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Imagine a large binary blob where all 0x00 bytes shall be replaced by   1 x30  and all 0x01 bytes by   1 x31  because the transport protocol allows no  0-bytes   In cases where    the replacing and the to-replaced string have different lengths  there are many occurences of the to-replaced string within the source string and the source string is large    the provided solutions cannot be applied  because they replace only single characters  or have a performance problem  because they would call string  replace several times which generates copies of the size of the blob over and over   I do not know the boost solution  maybe it is OK from that perspective   This one walks along all occurrences in the source string and builds the new string piece by piece once   void replaceAll std  string amp  source  const std  string amp  from  const std  string amp  to        std  string newString      newString reserve source length         avoids a few memory allocations      std  string  size type lastPos   0      std  string  size type findPos       while std  string  npos     findPos   source find from  lastPos                  newString append source  lastPos  findPos - lastPos           newString    to          lastPos   findPos   from length                  Care for the rest after last occurrence     newString    source substr lastPos        source swap newString

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If you re looking to replace more than a single character  and are dealing only with std  string  then this snippet would work  replacing sNeedle in sHaystack with sReplace  and sNeedle and sReplace do not need to be the same size  This routine uses the while loop to replace all occurrences  rather than just the first one found from left to right   while sHaystack find sNeedle     std  string  npos      sHaystack replace sHaystack find sNeedle  sNeedle size   sReplace

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If you re willing to use std  strings  you can use this sample-app s strsub function as-is  or update it if you want it to take a different type or set of parameters to achieve roughly the same goal  Basically  it uses the properties and functionalities of std  string to quickly erase the matching set of characters  and insert the desired characters directly within the std  string  Every time it does this replacement operation  the offset updates if it can still find matching chars to replace  and if it can t due to nothing more to replace  it returns the string in its state from the last update    include  lt iostream gt   include  lt string gt   std  string strsub std  string stringToModify                     std  string charsToReplace                     std  string replacementChars    int main         std  string silly typos    annoiiyyyng syyyllii tiipos         std  cout  lt  lt   Look at these    lt  lt  silly typos  lt  lt  std  endl      silly typos   strsub silly typos   yyy    i        std  cout  lt  lt   After a little elbow-grease  a few less    lt  lt  silly typos  lt  lt  std  endl      silly typos   strsub silly typos   ii    y         std  cout  lt  lt   There  no more    lt  lt  silly typos  lt  lt  std  endl      return 0     std  string strsub std  string stringToModify                     std  string charsToReplace                     std  string replacementChars        std  string this string   stringToModify       std  size t this occurrence   this string find charsToReplace       while  this occurrence    std  string  npos                this string erase this occurrence  charsToReplace size             this string insert this occurrence  replacementChars           this occurrence   this string find charsToReplace                                             this occurrence   replacementChars size                return this string      If you don t want to rely on using std  strings as your parameters so you can pass in C-style strings instead  you can see the updated sample below    include  lt iostream gt   include  lt string gt   std  string strsub const char   stringToModify                     const char   charsToReplace                     const char   replacementChars                     uint64 t sizeOfCharsToReplace                     uint64 t sizeOfReplacementChars    int main         std  string silly typos    annoiiyyyng syyyllii tiipos         std  cout  lt  lt   Look at these    lt  lt  silly typos  lt  lt  std  endl      silly typos   strsub silly typos c str     yyy    i   3  1       std  cout  lt  lt   After a little elbow-grease  a few less    lt  lt  silly typos  lt  lt  std  endl      silly typos   strsub silly typos c str     ii    y   2  1        std  cout  lt  lt   There  no more    lt  lt  silly typos  lt  lt  std  endl      return 0     std  string strsub const char   stringToModify                     const char   charsToReplace                     const char   replacementChars                     uint64 t sizeOfCharsToReplace                     uint64 t sizeOfReplacementChars        std  string this string   stringToModify       std  size t this occurrence   this string find charsToReplace       while  this occurrence    std  string  npos                this string erase this occurrence  sizeOfCharsToReplace           this string insert this occurrence  replacementChars           this occurrence   this string find charsToReplace              this occurrence   sizeOfReplacementChars              return this string

User · Answer

A simple find and replace for a single character would go something like   s replace s find  x    1   y    To do this for the whole string  the easy thing to do would be to loop until your s find starts returning npos  I suppose you could also catch range error to exit the loop  but that s kinda ugly

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I thought I d toss in the boost solution as well    include  lt boost algorithm string replace hpp gt      in place std  string in place    blah blah   boost  replace all in place                 copy const std  string input    blah blah   std  string output   boost  replace all copy input

[c++] How to replace all occurrences of a character in string?

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