How do you insert a string within a dynamic target using realloc()? - c

Question

I'm struggling with memory allocation. I wanted to input a string into another and I made two functions that stop working at the same place - realloc. These functions are very similar. In first one I copy char by char into a temporary string and when I try to copy temporary string to the first one is the place where I get errors. In the second function I copy the end of first string (from the given position) to a temporary string, reallocate the first string (this is where I get errors) and remove everything in i from the given position. Then I append second string and temporary to a first string. Here is my code.
First function:
// str2 - is a string that I want to input in first string(str)
// at certain position (pos)
void ins (char **str, char *str2, int pos)
{
// lenght of first and second strings
int len = strlen(str[0]),
len2 = strlen(str2),
i, j, l = 0;
// creating temporary string
char *s = (char *) malloc ((len + len2) * sizeof(char));
// copying first part of first string
for (i = 0; i < pos; i++)
s[i] = str[0][i];
// copying second string
for (j = 0; j < len2; j++)
s[i + j] = str2[j];
// copying second part of first string
for (int k = pos; k < len; k++)
{
s[i + j + l] = str[0][k];
l++;
}
// reallocating additional space for second string
// and copying temporary string to first string
str[0] = (char *) realloc (str[0], (len + len2) * sizeof(char));
strcpy(str[0], s);
free(s);
s = NULL;
}
Second function:
void ins2 (char **str,char *str2, int pos)
{
// lenght of first and second string
int len = strlen(str[0]),
len2 = strlen(str2);
// creating a temporary string and copying
// from the given position
char *s = (char *) malloc ((len - pos) * sizeof(char));
strcpy(s, str[0] + pos);
// reallocating space for string that will be added
// deleting part of it from the given position
str[0] = (char *) realloc(str[0], (len + len2) * sizeof(char));
str[0][pos] = '\0';
// adding second string and temporary string
strcat(str[0], str2);
strcat(str[0], s);
// be free, temporary string
free(s);
s = NULL;
}

Answer 1

If you're doing what I think you're trying to do, you need one realloc() for this, assuming the incoming string is indeed already dynamically allocated (it better be):
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
void ins (char **str, const char *str2, size_t pos)
{
// lenght of first and second strings
size_t len = strlen(*str);
size_t len2 = strlen(str2);
// reallocate new string
char *tmp = realloc(*str, len + len2 + 1);
if (tmp != NULL)
{
*str = tmp;
memmove(tmp+pos+len2, tmp+pos, len-pos);
memcpy(tmp+pos, str2, len2);
tmp[len+len2] = 0;
}
}
int main()
{
char *str = strdup("A simple string");
char s2[] = "inserted ";
printf("%s\n", str);
ins(&str, s2, 9);
printf("%s\n", str);
free(str);
return 0;
}
Output
A simple string
A simple inserted string
How It Works
The passed-in strings are both sent through strlen() to obtain their lengths. Once we have those we know how large the resulting buffer needs to be.
Once we realloc() the buffer, the original content is preserved, but we need to (possibly) shift content of the first string to open a hole for the second string. That shift, if done, may require overlapped memory be moved (as it does in the sample). For such memory copying, memmove() is used. Unlike memcpy(), the memmove() library function supports copying where the source and destination regions may overlap.
Once the hole is made, we memcpy() the second string into position. There is no need for strcpy() since we already know the length.
We finish by tacking the last slot to a terminating 0, thereby finishing the null-terminated string and completing the operation
Note I made no affordances at all for this regarding someone passing an invalid pos (out of range), NULL strings, optimizing to nothing if str2 is empty (or NULL), etc. That cleanup I leave to you, but I hope the idea of how this can be done is clear.