Question

This may sound a bit stupid but this is how I am getting it and I have googled it but no relevant solution was found.

I always thought that in C, strings was terminated by '\0' 

is a single character. (hence we used single quotes)

Also, we are able to do something like,

while ( a[i]!='\0' )
do blah blah


which suggests that '\0'

is a single character and ideally should be stored in a single position.

But when I declare an array like this:

char a[3];

and when I try to put some value say “hi” in it.

Then

printing a[0] gives "h"
printing a[1] gives empty space  
printing a[2] gives 0

which is suggestive that \ was stored at position 1 and ‘0’ at position 2. and the whole thing ‘\0’ was not stored together despite us using it as a single character.

Why is it so? Can anyone shed some clarity on the same?

Thanks!

EDIT:

#include <stdio.h>
void main()
{
int i=0;
char a[2];
fgets(a,sizeof(a),stdin); // Here I input "Hi".



while(a[i]!='\0')
    {
    printf("%c",a[i]);
    fflush(stdin);
    i=i+1;
    }
}

Answer 1

fgets takes the size of the buffer as the second argument and reads that minus one characters, then makes the last character a \0. Because your array is sized 2 (which is too small for the string hi, by the way), fgets reads one character and makes the last character a \0, so you have an array that holds h\0.

To get it working, make your array sized 3; one for the h, one for the i, and one for the \0:

#include <stdio.h>

void main() {
    int i = 0;
    char a[3];
    fgets(a, sizeof(a), stdin); // Here I input "Hi".

    while(a[i] != '\0') {
        printf("%c",a[i]);
        fflush(stdin);
        i=i+1;
    }
}