Replacing character in a string [duplicate]

Answer 1

#include <stdio.h>
#include <string.h>

int main(void)
{
    char temp[20];
    char s[20] = "aaabaa";
    int i, j;
    for (i=0; s[i]!= '\0'; i++)
    {
        if (s[i] == 'b')
        {
            strcpy(temp,s[i+1]); //copy rest of the string in this case 'aa'
            for (j=0; j<5; j++)
            {
                s[i+j] = 'c';
            }
            s[i+j] = '\0';   // here we get s = "aaaccccc"
            strcat(s,temp); // concat rest of the string (temp = "aa") after job is done. 
                           //  to this point s becomes s = "aaacccccaa"
        }
    }
    printf("%s\n", s); //s = "aaacccccaa". 
}

here we are using a buffer (temp) to store the rest of the string after our to be replaced character. after the replacement is done we append it to the end.

so we get s = "aaacccccaa"

Answer 2

It is necessary to declare a second char array. In below code it just copies content of array s to s1 when condition fails.

#include <stdio.h>
#include <string.h>
int main(void)
{
  char s[20] = "aaabaa";
  char s1[1024];
  int i, j, n;
  for (i=0, n = 0; s[i]!= '\0'; i++)
  {
    if (s[i] == 'b')
    {
        for (j=0; j<5; j++)
        {
            s1[n] = 'c';
            n++;
        }
    }
    else
    {
        s1[n] = s[i];
        n++;
    }
}
s1[n] = '\0';
printf("%s\n", s1);
}

Answer 3

Use this function :

char *replace(char *st, char *orig, char *repl) {
  static char buffer[4096];
  char *ch;
  if (!(ch = strstr(st, orig)))
   return st;
  strncpy(buffer, st, ch-st);  
  buffer[ch-st] = 0;
  sprintf(buffer+(ch-st), "%s%s", repl, ch+strlen(orig));
  return buffer;
  }

for your case : printf("%s\n", replace(s,"b","ccccc"));

Answer 4

Your problem is that you replace the "ccccc" into the original string thus overwriting the remaining characters after what you wish to replace... You should copy into a new string and keep track of two indices - one in each.

And be happy that you declared char s[20] larger than the size of your original string plus the replace values, as otherwise you'd have created a buffer overflow vulnerability in your critical login system :-)

Cheers,

Answer 5

Well, if you're going to dynamically allocate the array, you will probably have to allocate a second array. This is necessary because your string s only has a fixed amount of memory allocated.

So, instead of tryig to overwrite the characters in your for loop, I would suggest incrementing a counter that told you how big your new array has to be. Your counter should start off as the size of your original string and increment by 4 each time an instance of 'b' is found. You should then be able to write a function that appropriately copies the modified string over to a new char buffer of size[counter], inserting 5 c's every time a 'b' is being found.

Answer 6

If you want to do this in general, without worrying about trying to size your buffers, you should malloc a new string just large enough to hold the result:

/* return a new string with every instance of ch replaced by repl */
char *replace(const char *s, char ch, const char *repl) {
    int count = 0;
    const char *t;
    for(t=s; *t; t++)
        count += (*t == ch);

    size_t rlen = strlen(repl);
    char *res = malloc(strlen(s) + (rlen-1)*count + 1);
    char *ptr = res;
    for(t=s; *t; t++) {
        if(*t == ch) {
            memcpy(ptr, repl, rlen);
            ptr += rlen;
        } else {
            *ptr++ = *t;
        }
    }
    *ptr = 0;
    return res;
}

Usage:

int main() {
    char *s = replace("aaabaa", 'b', "ccccc");
    printf("%s\n", s);
    free(s);
    return 0;
}