There is a slight difference between char
and char*
. The first is a single character whereas the later is a pointer to char
(which can point to variable number of char
objects).
The %s
format specifier really expects a C-style string, which should not only be of type char*
but is also expected to be null-terminated (see C string handling). If you want to print a single character, then use %c
instead.
As for the program, assuming that what I think you want is what you want, try something like this:
#include <stdlib.h>
#include <stdio.h>
#include <assert.h>
static void strmycpy(char *dest, const char *src, size_t n) {
char c;
while (n-- > 0) {
c = *src++;
*dest++ = c;
if (c == '\0') {
while (n-- > 0)
*dest++ = '\0';
break;
}
}
}
int main(int argc, char *argv[]) {
size_t maxbytes;
char *stringb;
if (argc != 3 || !(maxbytes = atoll(argv[2]))) {
fprintf(
stderr,
"Usage: strmycpy <input string> <numberofbytes>.\n"
"Maxbytes has to be more than or equal to 1 and keep "
"in mind for the null byte (\\0).\n"
);
return EXIT_FAILURE;
}
assert(maxbytes > 0);
if (!(stringb = malloc(maxbytes))) {
fprintf(stderr, "Sorry, out of memory\n");
return EXIT_FAILURE;
}
strmycpy(stringb, argv[1], maxbytes);
printf("The copied string is: %.*s\n", (int)maxbytes, stringb);
free(stringb);
return EXIT_SUCCESS;
}
But frankly speaking, this is so fundamental that explaining might just result in writing a book on C. So you will be a lot better off if you just read one already written. For a list of good C books and resources, see The Definitive C Book Guide and List
Hope it helps. Good Luck!