C int memory storage. Least signinficant vs most significant bits?

Question

I'd expect the following combination of two uint8_t (0x00 and 0x01) into one uint16_t to give me a value of 0x0001, when I combine them consecutively in memory. Instead I obtain 0x0100 = 256, which I'm surprised of.

#include <stdio.h>
#include <stdint.h>

int main(void){

    uint8_t u1 = 0x00, u2 = 0x01;
    uint8_t ut[2] = {u1, u2};
    uint16_t *mem16 = (uint16_t*) ut;

    printf("mem16 = %d\n", *mem16);

    return 0;
}

Could anyone explain me what I've missed in my current understanding of C memory? Thank you! :-)

And then study [What is the strict aliasing rule?](https://stackoverflow.com/questions/98650/what-is-the-strict-aliasing-rule) — Lundin, Apr 21 '20 at 10:20
In x86 systems the 16 bit words (and all type of integers) are in LITTLE ENDIAN order byte. The byte 0 (the first) is the low order, the byte N (N=1 in this case) is the high order byte. — Sir Jo Black, Apr 21 '20 at 10:21
Thank you for all your answers and comments! I understand better now :-) — L. Arnaud, Apr 21 '20 at 11:43

score 2 · Accepted Answer · answered Apr 21 '20 at 10:40

2

It is called endianess.

Most system nowadays use little endian. In this system first is stored the least significant byte. So the 0x0100 is stored (assuming 2 bytes representation) as {0x00, 0x01} exactly as in your case

answered Apr 21 '20 at 10:40

0___________

60,014
4
34
74

score -2 · Answer 2 · answered Apr 21 '20 at 10:42

-2

ut[0] is inserted on LSB of mem16 , and ut[1] on MSB.

answered Apr 21 '20 at 10:42

omid.azmoon

1
1

3

This does not really explain what the OP is missing, does it? – the busybee Apr 21 '20 at 10:46

C int memory storage. Least signinficant vs most significant bits?

2 Answers2