Shift a character forward by three spaces in C - c

Question

Write a function that, using the strrot function,
modifies each string of the given array s out of n strings.
The function prototype is:
void rot_all(char *s[], int n);
Here's my code I only don't know how to do the last function.
char rot3(char c) // Function shifts a letter by three spaces (eg 'a' -> 'd')
{
if (c >= 'A' && c <= 'Z') return c += 3;
if (c >= 'a' && c <= 'z') return c += 3;
return c;
}
void strrot(char *str)
{
int d = strlen (str);
for (int i = 0; i < d; i++)
{
str[i] = rot3(str[i]);
}
}
void rot_all(char *s[], int n)
{
}
int main ()
{
char str[23];
scanf("%s",str);
strrot(str);
printf("%s",str);
return 0;
}

Answer 1

The assumption is that c is a either lower or upper case letter. Using an assert() to document this.
#include <assert.h>
#include <ctype.h>
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
char rot3(char c) {
assert(isalpha(c));
char offset = isupper(c) ? 'A' : 'a';
return offset + (c - offset + 3) % ('Z' - 'A' + 1);
}
void strrot(char *str) {
for (int i = 0; str[i]; i++) {
str[i] = rot3(str[i]);
}
}
void rot_all(char *s[], int n) {
for(int i = 0; i < n; i++) {
strrot(s[i]);
}
}
int main () {
char *strs[] = {
strdup("Hello"),
strdup("World")
};
rot_all(strs, sizeof(strs) / sizeof(*strs));
for(int i = 0; i < sizeof(strs) / sizeof(*strs); i++) {
printf("%s\n", strs[i]);
free(strs[i]);
}
return 0;
}
and the output is:
Khoor
Zruog