How to add a character in a string by checking the delimiter? - c

Question

I have a string in my program where in which it need to be altered with another string value before a "/".
Source String : qos-tree/output_rate
Target String : qos-tree-2/output_rate
#include <stdio.h>
#include <string.h>
void append(char* s, char c)
{
int len = strlen(s);
s[len] = c;
s[len+1] = '\0';
}
int main(void)
{
char str[256] = "qos-tree/output_rate";
char c = "a";
append(str, c);
printf("%s\n", str);
return 0;
}
This is what i have done so far,I think the logic is wrong here.Can anyone guide me to correct it?
Once the execution is completed the source string should have a "-2" before the "/"

Answer 1

void insert_before_ch(char *s, const char *ins, char c){
char *p = strchr(s, c);
if(p){
size_t len = strlen(ins);
memmove(p + len, p, strlen(p)+1);
memcpy(p, ins, len);
}
}
int main(void){
char str[256] = "qos-tree/output_rate";
insert_before_ch(str, "-2", '/');
printf("%s\n", str);
return 0;
}

Answer 2

In your attempt, you don't look for a slash and I do not see any "-2" anywhere.
Try this instead:
#include <stdio.h>
#include <string.h>
void append(char* s, char del, char* substring) {
char origin[256];
strcpy(origin, s);
int i = 0, j = 0, z = 0;
for(; origin[i]; ++i) {
if(origin[i] != del) {
s[j++] = origin[i];
} else {
for(; substring[z]; ++z) {
s[j++] = substring[z];
}
s[j++] = origin[i];
}
}
s[j] = '\0';
}
int main(void) {
char str[256] = "qos-tree/output_rate";
char del = '/';
char* substring = "-2";
append(str, del, substring);
printf("%s\n", str);
return 0;
}
The logic is that inside the function we use origin array to remember the actual contents of the array and then we copy from origin to s (which is the actual array of main()). If we find our delimiter del, then we copy the substring in that position and continuing with copying.
Note that the length of the array should be enough to store the resulted string. In this case, it is.

Answer 3

I think you should make your function work with dynamic allocation, because inserting characters into the string will make the resulting string larger, so this is my suggestion
#include <stdlib.h>
#include <stdio.h>
#include <string.h>
void insert(char **str, char chr, unsigned int position)
{
int length;
char *ptr;
if (str == NULL)
return;
length = strlen(*str);
if (position >= length)
return;
ptr = realloc(*str, 2 + length);
if (ptr == NULL)
return;
*str = ptr;
memmove(ptr + position + 1, ptr + position, length - position + 1);
ptr[position] = chr;
}
int main(void)
{
const char *source = "qos-tree/output_rate";
size_t length = strlen(source);
char *str = malloc(1 + length);
if (str == NULL)
return -1;
strcpy(str, source);
insert(&str, '-', 8);
insert(&str, '2', 9);
printf("%s\n", str);
free(str);
return 0;
}

Answer 4

first of all thist char c = "a" should be replace with this char c = 'a'; because c is a character not a string
as for your problem I didn't realy see the relation between what your code is doing with what you said you wanted to do , but here a piece of code to achieve what , I think , you want to do :
#include <string.h>
#include <stdio.h>
#include <stdlib.h>
void append(char* str , char c)
{
char firststr[60];
char therest[30];
strcpy(firststr , strtok(str , "/"));
strcpy(therest , strtok(NULL , "/"));
strcat(firststr , &c);
strcat(firststr , "/");
strcat(firststr , therest);
strcpy(str , firststr);
}
int main(void)
{
char str[60] = "qos-tree/output_rate";
char c = '2';
append(str , c);
printf("%s\n" , str);
}
there you go I think this is what you wanted to do you can modify the array sizes to fit your needs