What is the function to replace string in C?

Question

Given a (char *) string, I want to find all occurrence of a substring and replace it with an alternate string. I do not see any simple function that achieves this in <string.h>

Answer 1

You could build your own replace function using strstr to find the substrings and strncpy to copy in parts to a new buffer.

Unless what you want to replace_with is the same length as what you you want to replace, then it's probably best to use a new buffer to copy the new string to.

Answer 2

There isn't one.

You'd need to roll your own using something like strstr and strcat or strcpy.

Answer 3

This is not provided in the standard C library because, given only a char* you can't increase the memory allocated to the string if the replacement string is longer than the string being replaced.

You can do this using std::string more easily, but even there, no single function will do it for you.

Answer 4

As strings in C can not dynamically grow inplace substitution will generally not work. Therefore you need to allocate space for a new string that has enough room for your substitution and then copy the parts from the original plus the substitution into the new string. To copy the parts you would use strncpy.

Answer 5

If these string manipulations are critical part of your application I recommend getting a regular expression library for c.

Answer 6

The optimizer should eliminate most of the local variables. The tmp pointer is there to make sure strcpy doesn't have to walk the string to find the null. tmp points to the end of result after each call. (See Shlemiel the painter's algorithm for why strcpy can be annoying.)

Edit: didn't notice the "find all occurrences" the first time:

// You must free the result if result is non-NULL.
char *str_replace(char *orig, char *rep, char *with) {
    char *result; // the return string
    char *ins;    // the next insert point
    char *tmp;    // varies
    int len_rep;  // length of rep
    int len_with; // length of with
    int len_front; // distance between rep and end of last rep
    int count;    // number of replacements

    if (!orig)
        return NULL;
    if (!rep || !(len_rep = strlen(rep)))
        return NULL;
    if (!(ins = strstr(orig, rep))) 
        return NULL;
    if (!with)
        with = "";
    len_with = strlen(with);

    for (count = 0; tmp = strstr(ins, rep); ++count) {
        tmp += ins - tmp;
        ins = tmp;
    }

    // first time through the loop, all the variable are set correctly
    // from here on,
    //    tmp points to the end of the result string
    //    ins points to the next occurrence of rep in orig
    //    orig points to the remainder of orig after "end of rep"
    tmp = result = malloc(strlen(orig) + (len_with - len_rep) * count + 1);

    if (!result)
        return NULL;

    while (1) {
        len_front = ins - orig;
        tmp = strncpy(tmp, orig, len_front) + len_front;
        tmp = strcpy(tmp, with) + len_with;
        orig += len_front + len_rep; // move to next "end of rep"
        if (--count) break;
        ins = strstr(orig, rep);
    }
    strcpy(tmp, orig);
    return result;
}

Answer 7

Here's some sample code that does it.

#include <string.h>
#include <stdlib.h>

char * replace(
    char const * const original, 
    char const * const pattern, 
    char const * const replacement
) {
  size_t const replen = strlen(replacement);
  size_t const patlen = strlen(pattern);
  size_t const orilen = strlen(original);

  size_t patcnt = 0;
  const char * oriptr;
  const char * patloc;

  // find how many times the pattern occurs in the original string
  for (oriptr = original; patloc = strstr(oriptr, pattern); oriptr = patloc + patlen)
  {
    patcnt++;
  }

  {
    // allocate memory for the new string
    size_t const retlen = orilen + patcnt * (replen - patlen);
    char * const returned = (char *) malloc( sizeof(char) * (retlen + 1) );

    if (returned != NULL)
    {
      // copy the original string, 
      // replacing all the instances of the pattern
      char * retptr = returned;
      for (oriptr = original; patloc = strstr(oriptr, pattern); oriptr = patloc + patlen)
      {
        size_t const skplen = patloc - oriptr;
        // copy the section until the occurence of the pattern
        strncpy(retptr, oriptr, skplen);
        retptr += skplen;
        // copy the replacement 
        strncpy(retptr, replacement, replen);
        retptr += replen;
      }
      // copy the rest of the string.
      strcpy(retptr, oriptr);
    }
    return returned;
  }
}

#include <stdio.h>
int main(int argc, char * argv[])
{
  if (argc != 4)
  {
    fprintf(stderr,"usage: %s <original text> <pattern> <replacement>\n", argv[0]);
    exit(-1);
  }
  else
  {
    char * const newstr = replace(argv[1], argv[2], argv[3]);
    if (newstr)
    {
      printf("%s\n", newstr);
      free(newstr);
    }
    else
    {
      fprintf(stderr,"allocation error\n");
      exit(-2);
    }
  }
  return 0;
}

Answer 8

Welcome to the deep end of the pool---I see you've found your cinderblock :-)

You could roll your own, but get Dave Hanson's C Interfaces and Implementations. It includes a powerful string-processing library and many other useful modules. The software is free, and the book is worth buying.

Answer 9

http://www.planet-source-code.com/vb/scripts/ShowCode.asp?txtCodeId=10890&amp;lngWId=3