Add a char padding in C strings - arrays

Question

I am trying to implement this simple encryption method for a a list of numbers, the encryption is like this:
we add the 1st element of the list before every element of the entire list of numbers,
if we have:
char array = "356307042441013"
the first number is 3, that means we need to add it before every element of the list of numbers:
'33 35 36 33 30 37 30 34 32 34 34 31 30 31 33'
char result= "333536333037303432343431303133"
is there any function in C that will make the implementation easier ? because I tried doing it with shifting but couldn't get that result.

Answer 1

You can do following:
Step I: Allocate memory of double the size of input and + 1 to accommodate the null character to the result string.
Step II: Iterate through the input string and, in every iteration, first copy the input[0] character to current location of result string and, in very next location of result string, copy the current processing character of input string.
Step III: Once the loop exits, add null character at the end of result string.
[Take special care of empty string because it will have nothing to encrypt]
Implementation:
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char * encryption (const char * input) {
if ((!input) || (*input == '\0')) {
return NULL;
}
size_t len = strlen (input);
char * temp = malloc ((len * 2) + 1);
if (!temp) {
printf ("Failed to allocate memory\n");
exit (EXIT_FAILURE);
}
size_t i = 0, j = 0;
for (; input[j]; ++j) {
temp[i++] = input[0];
temp[i++] = input[j];
}
temp[i] = '\0';
return temp;
}
int main (void) {
char array[] = "356307042441013";
char * result = encryption (array);
if (result) {
printf ("result : %s\n", result);
// Free the dynamically allocated memory once done with it
free (result);
}
return 0;
}
Output:
# ./a.out
result : 333536333037303432343431303133

Answer 2

A cleaner, more concise solution:
#include <stdio.h>
#include <string.h>
int main()
{
char foo[] = "356307042441013";
char bar[2 * sizeof(foo) - 1];
char *src = foo, *dest = bar;
while (*src) {
*dest++ = foo[0];
*dest++ = *src++;
}
*dest = *src;
printf("foo=%s\n",foo);
printf("bar=%s\n",bar);
return 0;
}
It is better to use foo[] rather than hard-coding the length, because what if you want to change the string. If you use [] (empty brackets) the compiler will allocate exactly how many bytes you need (including the terminating null). Similarly, for bar we base the size on the size of foo by doubling it and subtracting 1 (since the terminating null does not need to be doubled).

Answer 3

I found a way to solve this:
int main() {
char foo[16] = "356307042441013";
char bar[2*16-1];
for (int i = 0; i < 16; i++) {
bar[2*i] = foo[i];
if (i != 16 - 1)
bar[2*i + 1] = foo[0];
}
char res[32]= "3";
strcat(res,bar);
res[30] = '\0';
printf("bar=%s\r\n",bar);
printf("begin=%s\r\n",res);
return 0;
}