Why does the following print what it does?

Question

typedef unsigned char byte;

unsigned int nines = 999;
byte * ptr = (byte *) &nines;


printf ("%x\n",nines);
printf ("%x\n",nines * 0x10);
printf ("%d\n",ptr[0]);
printf ("%d\n",ptr[1]);
printf ("%d\n",ptr[2]);
printf ("%d\n",ptr[3]);

Output:

3e7
3e70
231
3
0
0

I know the first two are just hexadecimal representations of 999 and 999*16. What do the remaining 4 mean? the ptr[0] to ptr[3]?

Answer 1

Most likely you are running this on a 32 bit LE system 999 in hex is:-
00 00 03 E7 - The way it would be stored in memory would be
E7 03 00 00 Hence:-

ptr[0] points to the byte containing E7 which is 231 in decimal
ptr[1] points to the byte containing 03 which is 3 in decimal
ptr[2] points to the byte containing 00 which is 0 in decimal
ptr[3] points to the byte containing 00 which is 0 in decimal

HTH!

Answer 2

I think that you will see clearly if you write:

typedef unsigned char byte;

main() {
unsigned int nines = 999;
byte * ptr = (byte *) &nines;

printf ("%x\n",nines);
printf ("%x\n",nines * 0x10);
printf ("%x\n",ptr[0]);
printf ("%x\n",ptr[1]);
printf ("%x\n",ptr[2]);
printf ("%x\n",ptr[3]);
printf ("%d\n",sizeof(unsigned int));
}

char is 8 bits, one byte, and int is 4 bytes (in my 64 bytes machine). In your machine the data is saved as little-endian so less significative byte is located first.