What happens when pointers are typecasted?

Question

During some trial and error, came across this. It is outputting 513 as the value, cant figure out why.

int a;
char *x;
x =(char *) &a;
a = 512;
x[0]=1;
x[1]=2;
printf("%d\n ",a);

`(2<<8)+1` equals 513 on a little endian machine – Jean-François Fabre Jul 22 '19 at 16:31 — Jean-François Fabre, Jul 22 '19 at 16:31

score 1 · Answer 1 · answered Jul 22 '19 at 16:33

On a typical x86 platform an int is four bytes wide and is stored little endian. Little endian means the bytes are stored in reverse, least significant to most significant. The number 512 is represented in memory as 2 × 2⁸, or:

{ 0x00, 0x02, 0x00, 0x00 }
// LSB              MSB
// 2^0  2^8   2^16  2^24

After the two assignments you have:

{ 0x01, 0x02, 0x00, 0x00 }
// LSB              MSB
// 2^0  2^8   2^16  2^24

Converting these four bytes back to decimal you have 1 × 2⁰ + 2 × 2⁸ = 1 + 512 = 513.

it also works if integers are 8 or 16 or whatever size, since it's little endian — Jean-François Fabre, Jul 22 '19 at 16:34

What happens when pointers are typecasted?

1 Answers1