Why the pattern of data appears in memory when I use `malloc()`?

I have a simple code:

#include <stdio.h>
#include <stdlib.h>

int main(void){
  char *str = (char *) malloc(4*sizeof(char));
  int i;
  for(i = 0; i < 64; i ++)
    printf("%d, %ld, %d, %c\n", i, (long) &(str[i]), (int) str[i], str[i]);
  return 0;
}

I allocate a memory into str using malloc() which is available to save 4 letters in str[0], ..., str[3]. I know that malloc() does not initialize its memory while calloc() does.

This program prints str[i] with i, address of str[i], value of str[i], letter of str[i], in order. (I use 64-bits Ubuntu, hence address is long type.)

As expected, addresses are quite different for every time I run the program. But I wonder that why str[24], str[25], and str[26] are -31, 15, 2, repectively, and other values are all 0 as you can see below:

(Note that without option -O0 gives same result.)

How can memory has same sequence (0,0,...,0,-31,15,2,0,0,...) even though only first four 0s in that sequence are allocated and others are out of care?