Type casting a 16 bit unsigned integer to an 8 bit unsigned integer pointer in C

Question

While playing around with pointers I came around something interesting.

I initialized a 16 bit unsigned int variable to the number 32771 and then assigned the address of that variable to an 8 bit unsigned int pointer.

Now 32771, in unsigned 16 bit form, has binary representation of 110000000000001. So when dereferencing the 8 bit pointer the first time, I expected it to print the value of 11000000, which is = 192 and after incrementing the pointer and then dereferencing the pointer again, is expected it to print the value 00000001, which is 128.

In actuality, for the first dereference, 3 was printed, which is what I would get if I read 11000000 from left to right and the second dereference printed 128.

int main(){
    __uint16_t a = 32771;
    __uint8_t *p = (__uint8_t *)&a;
    printf("%d", *p); //Expected 192 but actual output 3.
    ++p;
    printf("%d", *p); //Expected 1, but actual output 128
}

I know that bits are read from right to left, however in this case the bits are being read from left to right. Why?

Answer 1

32771 is 32768 plus 3. In binary, it's 1000000000000011.

If you split it into the most significant byte and the last significant byte, you get 128 and 3 because 128 * 256 + 3 = 32771.

So the bytes will be 3 and 128. There's no particular reason one should occur before the other. Your CPU can store the bytes that compose a multi-byte number in whatever order it wants to. Apparently, yours stores the least significant byte at a lower address than the most significant byte.