Weird output of an array in c programming

Question

I am writing a program in C to separate the hour and minutes using arrays. However, the program gives me some outputs which exceed the length of the defined array. Can someone explain to me? I want the two array only include hour and the rest of the time without (AM/PM).

#include <math.h>
#include <stdio.h>
#include <string.h>
#include <stdlib.h>
#include <assert.h>
#include <limits.h>
#include <stdbool.h>

int main() {
    char* s = (char *)malloc(10 * sizeof(char));
    s="11:22:33AM";

    printf("The time is %s\n",s);

    char Hour[2];
    char Minutes[6];
    int i;
    printf("The hour is %s\n",Hour);
    printf("The minute is %s\n",Minutes);

    for (i=0;i<2;i++){
        Hour[i]=s[i];
    }
   for (i=2;i<8;i++){
        Minutes[i-2]=s[i];
    }

    printf("%d\n",sizeof(Hour));
    printf("%d\n",sizeof(Minutes));

    printf("The hour is %s\n",Hour);
    printf("The minute is %s\n",Minutes);

}

Answer 1

The %s specifier is for null-terminated strings but your arrays are fixed-length character buffers. You should use the format-string specifier that sets the length of the output instead %.{length}s:

printf( "The hour is %.2s\n", Hour );

See the documentation for printf: http://www.cplusplus.com/reference/cstdio/printf/

.precision : this is the maximum number of characters to be printed. By default all characters are printed until the ending null character is encountered.

As your character-buffers do not contain terminal nulls '\0' there won't be a null character so the computer will keep on scanning memory until it does encounter a zero.

An alternative is to increase the size of the buffers and set their last character value to nulls.

Answer 2

You just need to add at the end of both arrays the terminate character to create a null-terminated string. So your code will become:

char Hour[3];
char Minutes[7];
.
.
.
Hour[2] = '\0';    //or  Hour[2] = 0;
Minutes[6] = '\0'; //or  Minutes[6] = 0;

Also don't forget at the end of the program to free the dynamically allocated memory free(s);

Answer 3

There are a number of approaches you can take. There is no need to allocate, as your initial time can be assigned directory to s as a string-literal. Then it is simply a matter of filling hours and minutes with strncpy to separate the time. No loops required. Also, sizeof hours or sizeof minutes return the total size of the arrays, not the length within. For that (and since you initialized the arrays to all zero providing default nul-termination when one-less than the total number of characters are filled -- you can use both as strings, e.g.

#include <stdio.h>
#include <string.h>

int main (void) {

    char *s = "11:22:33AM",     /* no allocation required */
        hour[3] = "",           /* initialize all arrays to zero */
        minutes[6] = "";

    printf ("The time is %s\n", s);

    /* use strncpy to fill both hour and minutes */
    printf ("The hour is %s\n", strncpy (hour, s, 2));
    printf ("The minute is %s\n", strncpy (minutes, s + 3, 5));

    printf ("%zu\n", strlen (hour));
    printf ("%zu\n", strlen (minutes));

    printf("The hour is %s\n", hour);
    printf("The minute is %s\n", minutes);

}

Example Use/Output

$ ./bin/splittime
The time is 11:22:33AM
The hour is 11
The minute is 22:33
2
5
The hour is 11
The minute is 22:33

Look things over and let me know if that is what you intended. If not, drop a comment and I'm happy to help further.