Pointers in C

Hey guys I'm having a little trouble with the following example

#include <stdio.h>
int main(void)
{
char s[] = "Melbourne City";
printf("%s\n", &s[2]);
return 0;
}

which prints "lbourne City" instead of the address of letter 'l'.
I do understand that the %s in the printf continues until a null terminator is hit, but not why &s[2] returns a character in the first place.
Thanks.

>>78
Here, you definitely want an unsigned value, as a negative sizeof return value would not make much sense. Again, as the size of the array is merely 7, it does not make an actual difference. If on a 16-bit-integer system you had an array larger than 32k (or on a 32-bit-integer system an array larger than 2G), the expression (int)sizeof(arr) would return a negative value - however, the %u specifier in the string literal would recast it to unsigned int anyway. To be totally proper, the return value of sizeof should be cast from size_t to unsigned int in the first place:

printf("\tsizeof(arr): %u\n", (unsigned int)sizeof(arr));
Then printf("\tsizeof(arr): %zu\n", sizeof(arr));