What is the function to replace string in C?

问题:

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>

回答1:

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.)

// 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 (the string to remove)
    int len_with; // length of with (the string to replace rep with)
    int len_front; // distance between rep and end of last rep
    int count;    // number of replacements

    // sanity checks and initialization
    if (!orig || !rep)
        return NULL;
    len_rep = strlen(rep);
    if (len_rep == 0)
        return NULL; // empty rep causes infinite loop during count
    if (!with)
        with = \"\";
    len_with = strlen(with);

    // count the number of replacements needed
    ins = orig;
    for (count = 0; tmp = strstr(ins, rep); ++count) {
        ins = tmp + len_rep;
    }

    tmp = result = malloc(strlen(orig) + (len_with - len_rep) * count + 1);

    if (!result)
        return NULL;

    // 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\"
    while (count--) {
        ins = strstr(orig, rep);
        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\"
    }
    strcpy(tmp, orig);
    return result;
}

回答2:

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.

回答3:

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.

回答4:

There isn\'t one.

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

回答5:

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.

回答6:

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;
}

回答7:

// Here is the code for unicode strings!


int mystrstr(wchar_t *txt1,wchar_t *txt2)
{
    wchar_t *posstr=wcsstr(txt1,txt2);
    if(posstr!=NULL)
    {
        return (posstr-txt1);
    }else
    {
        return -1;
    }
}

// assume: supplied buff is enough to hold generated text
void StringReplace(wchar_t *buff,wchar_t *txt1,wchar_t *txt2)
{
    wchar_t *tmp;
    wchar_t *nextStr;
    int pos;

    tmp=wcsdup(buff);

    pos=mystrstr(tmp,txt1);
    if(pos!=-1)
    {
        buff[0]=0;
        wcsncpy(buff,tmp,pos);
        buff[pos]=0;

        wcscat(buff,txt2);

        nextStr=tmp+pos+wcslen(txt1);

        while(wcslen(nextStr)!=0)
        {
            pos=mystrstr(nextStr,txt1);

            if(pos==-1)
            {
                wcscat(buff,nextStr);
                break;
            }

            wcsncat(buff,nextStr,pos);
            wcscat(buff,txt2);

            nextStr=nextStr+pos+wcslen(txt1);   
        }
    }

    free(tmp);
}

回答8:

The repl_str() function on creativeandcritical.net is fast and reliable. Also included on that page is a wide string variant, repl_wcs(), which can be used with Unicode strings including those encoded in UTF-8, through helper functions - demo code is linked from the page. Belated full disclosure: I am the author of that page and the functions on it.

回答9:

i find most of the proposed functions hard to understand - so i came up with this:

static char *dull_replace(const char *in, const char *pattern, const char *by)
{
    size_t outsize = strlen(in) + 1;
    // TODO maybe avoid reallocing by counting the non-overlapping occurences of pattern
    char *res = malloc(outsize);
    // use this to iterate over the output
    size_t resoffset = 0;

    char *needle;
    while (needle = strstr(in, pattern)) {
        // copy everything up to the pattern
        memcpy(res + resoffset, in, needle - in);
        resoffset += needle - in;

        // skip the pattern in the input-string
        in = needle + strlen(pattern);

        // adjust space for replacement
        outsize = outsize - strlen(pattern) + strlen(by);
        res = realloc(res, outsize);

        // copy the pattern
        memcpy(res + resoffset, by, strlen(by));
        resoffset += strlen(by);
    }

    // copy the remaining input
    strcpy(res + resoffset, in);

    return res;
}

output must be free\'d

回答10:

Here is the one that I created based on these requirements:

1. Replace the pattern regardless of whether is was long or shorter.

2. Not use any malloc (explicit or implicit) to intrinsically avoid memory leaks.

3. Replace any number of occurrences of pattern.

4. Tolerate the replace string having a substring equal to the search string.

5. Does not have to check that the Line array is sufficient in size to hold the replacement. e.g. This does not work unless the caller knows that line is of sufficient size to hold the new string.

/* returns number of strings replaced.
*/
int replacestr(char *line, const char *search, const char *replace)
{
   int count;
   char *sp; // start of pattern

   //printf(\"replacestr(%s, %s, %s)\\n\", line, search, replace);
   if ((sp = strstr(line, search)) == NULL) {
      return(0);
   }
   count = 1;
   int sLen = strlen(search);
   int rLen = strlen(replace);
   if (sLen > rLen) {
      // move from right to left
      char *src = sp + sLen;
      char *dst = sp + rLen;
      while((*dst = *src) != \'\\0\') { dst++; src++; }
   } else if (sLen < rLen) {
      // move from left to right
      int tLen = strlen(sp) - sLen;
      char *stop = sp + rLen;
      char *src = sp + sLen + tLen;
      char *dst = sp + rLen + tLen;
      while(dst >= stop) { *dst = *src; dst--; src--; }
   }
   memcpy(sp, replace, rLen);

   count += replacestr(sp + rLen, search, replace);

   return(count);
}

Any suggestions for improving this code are cheerfully accepted. Just post the comment and I will test it.

回答11:

/*замена символа в строке*/
char* replace_char(char* str, char in, char out) {
    char * p = str;

    while(p != \'\\0\') {
        if(*p == in)
            *p == out;
        ++p;
    }

    return str;
}

回答12:

You can use this function (the comments explain how it works):

void strreplace(char *string, const char *find, const char *replaceWith){
    if(strstr(string, replaceWith) != NULL){
        char *temporaryString = malloc(strlen(strstr(string, find) + strlen(find)) + 1);
        strcpy(temporaryString, strstr(string, find) + strlen(find));    //Create a string with what\'s after the replaced part
        *strstr(string, find) = \'\\0\';    //Take away the part to replace and the part after it in the initial string
        strcat(string, replaceWith);    //Concat the first part of the string with the part to replace with
        strcat(string, temporaryString);    //Concat the first part of the string with the part after the replaced part
        free(temporaryString);    //Free the memory to avoid memory leaks
    }
}

回答13:

a fix to fann95\'s response, using in-place modification of the string, and assuming the buffer pointed to by line is large enough to hold the resulting string.

static void replacestr(char *line, const char *search, const char *replace)
{
     char *sp;

     if ((sp = strstr(line, search)) == NULL) {
         return;
     }
     int search_len = strlen(search);
     int replace_len = strlen(replace);
     int tail_len = strlen(sp+search_len);

     memmove(sp+replace_len,sp+search_len,tail_len+1);
     memcpy(sp, replace, replace_len);
}

回答14:

There you go....this is the function to replace every occurance of char x with char y within character string str

char *zStrrep(char *str, char x, char y){
    char *tmp=str;
    while(*tmp)
        if(*tmp == x)
            *tmp++ = y; /* assign first, then incement */
        else
            *tmp++;

    *tmp=\'\\0\';
    return str;
}

An example usage could be

  Exmaple Usage
        char s[]=\"this is a trial string to test the function.\";
        char x=\' \', y=\'_\';
        printf(\"%s\\n\",zStrrep(s,x,y));

  Example Output
        this_is_a_trial_string_to_test_the_function.

The function is from a string library I maintain on Github, you are more than welcome to have a look at other available functions or even contribute to the code :)

https://github.com/fnoyanisi/zString

EDIT: @siride is right, the function above replaces chars only. Just wrote this one, which replaces character strings.

#include <stdio.h>
#include <stdlib.h>

/* replace every occurance of string x with string y */
char *zstring_replace_str(char *str, const char *x, const char *y){
    char *tmp_str = str, *tmp_x = x, *dummy_ptr = tmp_x, *tmp_y = y;
    int len_str=0, len_y=0, len_x=0;

    /* string length */
    for(; *tmp_y; ++len_y, ++tmp_y)
        ;

    for(; *tmp_str; ++len_str, ++tmp_str)
        ;

    for(; *tmp_x; ++len_x, ++tmp_x)
        ;

    /* Bounds check */
    if (len_y >= len_str)
        return str;

    /* reset tmp pointers */
    tmp_y = y;
    tmp_x = x;

    for (tmp_str = str ; *tmp_str; ++tmp_str)
        if(*tmp_str == *tmp_x) {
            /* save tmp_str */
            for (dummy_ptr=tmp_str; *dummy_ptr == *tmp_x; ++tmp_x, ++dummy_ptr)
                if (*(tmp_x+1) == \'\\0\' && ((dummy_ptr-str+len_y) < len_str)){
                /* Reached end of x, we got something to replace then!
                * Copy y only if there is enough room for it
                */
                    for(tmp_y=y; *tmp_y; ++tmp_y, ++tmp_str)
                        *tmp_str = *tmp_y;
            }
        /* reset tmp_x */
        tmp_x = x;
        }

    return str;
}

int main()
{
    char s[]=\"Free software is a matter of liberty, not price.\\n\"
             \"To understand the concept, you should think of \'free\' \\n\"
             \"as in \'free speech\', not as in \'free beer\'\";

    printf(\"%s\\n\\n\",s);
    printf(\"%s\\n\",zstring_replace_str(s,\"ree\",\"XYZ\"));
    return 0;
}

And below is the output

Free software is a matter of liberty, not price.
To understand the concept, you should think of \'free\' 
as in \'free speech\', not as in \'free beer\'

FXYZ software is a matter of liberty, not price.
To understand the concept, you should think of \'fXYZ\' 
as in \'fXYZ speech\', not as in \'fXYZ beer\'

回答15:

DWORD ReplaceString(__inout PCHAR source, __in DWORD dwSourceLen, __in const char* pszTextToReplace, __in const char* pszReplaceWith)
{
    DWORD dwRC = NO_ERROR;
    PCHAR foundSeq = NULL;
    PCHAR restOfString = NULL;
    PCHAR searchStart = source;
    size_t szReplStrcLen = strlen(pszReplaceWith), szRestOfStringLen = 0, sztextToReplaceLen = strlen(pszTextToReplace), remainingSpace = 0, dwSpaceRequired = 0;
    if (strcmp(pszTextToReplace, \"\") == 0)
        dwRC = ERROR_INVALID_PARAMETER;
    else if (strcmp(pszTextToReplace, pszReplaceWith) != 0)
    {
        do
        {
            foundSeq = strstr(searchStart, pszTextToReplace);
            if (foundSeq)
            {
                szRestOfStringLen = (strlen(foundSeq) - sztextToReplaceLen) + 1;
                remainingSpace = dwSourceLen - (foundSeq - source);
                dwSpaceRequired = szReplStrcLen + (szRestOfStringLen);
                if (dwSpaceRequired > remainingSpace)
                {
                    dwRC = ERROR_MORE_DATA;
                }

                else
                {
                    restOfString = CMNUTIL_calloc(szRestOfStringLen, sizeof(CHAR));
                    strcpy_s(restOfString, szRestOfStringLen, foundSeq + sztextToReplaceLen);

                    strcpy_s(foundSeq, remainingSpace, pszReplaceWith);
                    strcat_s(foundSeq, remainingSpace, restOfString);
                }

                CMNUTIL_free(restOfString);
                searchStart = foundSeq + szReplStrcLen; //search in the remaining str. (avoid loops when replWith contains textToRepl 
            }
        } while (foundSeq && dwRC == NO_ERROR);
    }
    return dwRC;
}

回答16:

char *replace(const char*instring, const char *old_part, const char *new_part)
{

#ifndef EXPECTED_REPLACEMENTS
    #define EXPECTED_REPLACEMENTS 100
#endif

    if(!instring || !old_part || !new_part)
    {
        return (char*)NULL;
    }

    size_t instring_len=strlen(instring);
    size_t new_len=strlen(new_part);
    size_t old_len=strlen(old_part);
    if(instring_len<old_len || old_len==0)
    {
        return (char*)NULL;
    }

    const char *in=instring;
    const char *found=NULL;
    size_t count=0;
    size_t out=0;
    size_t ax=0;
    char *outstring=NULL;

    if(new_len> old_len )
    {
        size_t Diff=EXPECTED_REPLACEMENTS*(new_len-old_len);
        size_t outstring_len=instring_len + Diff;
        outstring =(char*) malloc(outstring_len); 
        if(!outstring){
            return (char*)NULL;
        }
        while((found = strstr(in, old_part))!=NULL)
        {
            if(count==EXPECTED_REPLACEMENTS)
            {
                outstring_len+=Diff;
                if((outstring=realloc(outstring,outstring_len))==NULL)
                {
                     return (char*)NULL;
                }
                count=0;
            }
            ax=found-in;
            strncpy(outstring+out,in,ax);
            out+=ax;
            strncpy(outstring+out,new_part,new_len);
            out+=new_len;
            in=found+old_len;
            count++;
        }
    }
    else
    {
        outstring =(char*) malloc(instring_len);
        if(!outstring){
            return (char*)NULL;
        }
        while((found = strstr(in, old_part))!=NULL)
        {
            ax=found-in;
            strncpy(outstring+out,in,ax);
            out+=ax;
            strncpy(outstring+out,new_part,new_len);
            out+=new_len;
            in=found+old_len;
        }
    }
    ax=(instring+instring_len)-in;
    strncpy(outstring+out,in,ax);
    out+=ax;
    outstring[out]=\'\\0\';

    return outstring;
}

回答17:

This function only works if ur string has extra space for new length

void replace_str(char *str,char *org,char *rep)
{
    char *ToRep = strstr(str,org);
    char *Rest = (char*)malloc(strlen(ToRep));
    strcpy(Rest,((ToRep)+strlen(org)));

    strcpy(ToRep,rep);
    strcat(ToRep,Rest);

    free(Rest);
}

This only replaces First occurrence

回答18:

// Replace every occurence of a in str with b

void strrepl(char *str, const char *a, const char *b) {
    for (char *cursor = str; (cursor = strstr(cursor, a)) != NULL;){
        memmove(cursor + strlen(b), cursor + strlen(a), strlen(cursor) - strlen(a) + 1);
        for (int i = 0; b[i] != \'\\0\'; i++)
            cursor[i] = b[i];
        cursor += strlen(b);
    }
}

This is possibly the best you could do:

• just 8 lines
• just 1 temp pointer
• in-place modification
• works with arbitrary sized a and b char*
• works for deletion

if you don\'t want in-place modification, or you have to work with a const char*, this is another version of the same function that returns a new char* with modifications done. (don\'t forget to free the return value as strdup() is doing a malloc behind the scene)

char* astrrepl(const char *str, const char *a, const char *b) {
    char* strd = strdup(str);
    for (char *cursor = strd; (cursor = strstr(cursor, a)) != NULL;){
        memmove(cursor + strlen(b), cursor + strlen(a), strlen(cursor) - strlen(a) + 1);
        for (int i = 0; b[i] != \'\\0\'; i++)
            cursor[i] = b[i];
        cursor += strlen(b);
    }
    return strd;
}

Sample code :

int main(int argc, char const *argv[])
{
    char *a = strdup(\"Hello This Is Jack\");
    char *b = astrrepl(a, \" \", \"~~~\");
    printf(\"Out of place modification : %s\\n\", b);

    strrepl(a, \"Jack\", \"Fourchette\");
    printf(\"In place modification     : %s\\n\", a);
    return 0;
}

Output:

Out of place modification : Hello~~~This~~~Is~~~Jack
In place modification     : Hello This Is Fourchette