Replacing character in a string [duplicate] - c

Question

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
What is the function to replace string in C?
I am trying to replace a certain character in my string with multiple characters. Here is an example of what I am trying to do.
Say I have the string "aaabaa"
I want to replace all occurrences of the character "b" with 5 "c"s.
So when I am done, "aaabaa" becomes "aaacccccaa"
I have written the following code:
#include <stdio.h>
#include <string.h>
int main(void)
{
char s[20] = "aaabaa";
int i, j;
for (i=0; s[i]!= '\0'; i++)
{
if (s[i] == 'b')
{
for (j=0; j<5; j++)
{
s[i+j] = 'c';
}
}
}
printf("%s\n", s);
}
My output from this function is "aaaccccc". It appears that it just overwrites the last two a's with the c's. Is there any way I would have it so that these last couple of a's dont get overwritten?

Answer 1

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;
}

Answer 2

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 3

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 4

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

Answer 5

#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 6

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 7

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"));