cast a char into a char* in C - c

Question

I have a char array s[11]="0123456789"; and I want to be able to take each digit s[i](using a for loop) and cast it somehow to a char*(I need a char* specifically because I need to use strncat on some other string)
I've been trying to do this for the past 4 hours and I couldn't get anything done.

Answer 1

Unless you're willing to temporarily modify s, or copy the character somewhere else... You're going to have a hard time.
Modify s temporarily.
int main() {
char s[11] = "0123456789";
for (int i = 0; i < strlen(s); i++) {
// Modify s in place.
const char tmp = s[i+1];
s[i+1] = '\0';
char *substr = &s[i];
// Do something with substr.
printf("%s\n", substr);
// Fix s
s[i+1] = tmp;
}
}
Or copy s[i] to a new string.
int main() {
char s[11] = "0123456789";
for (int i = 0; i < strlen(s); i++) {
// Create a temporary string
char substr[2];
substr[0] = s[i];
substr[1] = '\0';
// Do something with substr.
printf("%s\n", substr);
}
}
Or maybe just don't use strncat?
Assuming you're actually using strncat or similar... Those functions are pretty trivial, especially if you are appending only a single character. You might just create your own version.
void append_character(char *buffer, int buffer_size, char new_character) {
int length = strlen(buffer);
if (length + 2 < buffer_size) {
buffer[length] = new_character;
buffer[length+1] = '\0';
} else {
// No space for an additional character, drop it like strncat would.
}
}
Or you could do it as a simple wrapper around strncat:
void append_character(char *buffer, int buffer_size, char new_character) {
char new_string[2] = { new_character, '\0' };
strncat(buffer, buffer_size, new_string);
}

Answer 2

What you are requesting does not make any sense. A char is a small number. A pointer is the address of some memory. Converting a char to a pointer won't give you a valid pointer, but a pointer that will crash as soon as it is used.
For strncat, you need an array of characters containing a C string. You could create a string by writing char array[2]; (now you have an array with space for two characters), then array[0] = whateverchar; array[1] = 0; and now you have a C string with space for exactly one char and one trailing zero byte.

Answer 3

Yet another idea:
#include <stdio.h>
#include <string.h>
int main(void)
{
char src[11]="0123456789";
char dest[50]="abc";
char* p = src;
int i = 0;
int len = strlen(src);
for (i = 0; i < len; i++)
{
strncat(dest,p,1);
p++;
}
printf("src: %s\n", src);
printf("dest: %s\n", dest);
return 0;
}
Compiled with gcc under Ubuntu:
$ gcc hello_str.c -o hello_str
Output:
$ ./hello_str
src: 0123456789
dest: abc0123456789