Pointer Typecasting: int to short in C

Asked May 22 '18 at 18:40

Active May 22 '18 at 18:59

Viewed 78 times

I expect the output to be 131200 (2^17 + 2^ 7) rather than 2. Makes sense in theory, but it is not working. The bit pattern of arr[3] after the pointer typecasting line would be something like - [00000000 00000010 00000000 10000000]. (Please tell me if I am wrong somewhere.)

#include <stdio.h>

int main(void){
    int arr[5];
    arr[3] = 128;
    ((short *) arr)[6] = 2;
    printf("%i\n", arr[3]);
}

edited May 22 '18 at 18:59

asked May 22 '18 at 18:40

Harry Sandhu

And if you do `printf("short: %d, int: %d\r\n", (int)sizeof(short), (int)sizeof(int));`? – unalignedmemoryaccess May 22 '18 at 18:45
2

`((short *) arr)[6] = 2;` is [a strict aliasing violation](https://stackoverflow.com/questions/98650/what-is-the-strict-aliasing-rule), undefined behavior, and a good way to get a `SIGBUS` or `SIGSEGV` on hardware that has alignment restrictions. – Andrew Henle May 22 '18 at 18:45
Why expect `2^9` and not `2^16 * 2` as `short` is commonly 16 bit. – chux - Reinstate Monica May 22 '18 at 18:52
2

Have you tried `((short *) arr)[6 + 1] = 2;`? – chux - Reinstate Monica May 22 '18 at 18:52
yea, that works! – Harry Sandhu May 22 '18 at 19:09
This also won't work on a big-endian machine. – sudo May 22 '18 at 19:14
but why at an offset of 7 from base address? @chux – Harry Sandhu May 22 '18 at 19:15
1

Make a diagram on paper, and appreciate the endianness. – Weather Vane May 22 '18 at 19:19
@sudo chux comment is little-endian, but the code is for big-endian, so it overwrites the earlier value. – Weather Vane May 22 '18 at 19:20
@HarrySandhu, Research [endian](https://en.wikipedia.org/wiki/Endianness) – chux - Reinstate Monica May 22 '18 at 19:21
Got it, Thanks everyone! – Harry Sandhu May 22 '18 at 19:26

Pointer Typecasting: int to short in C

0 Answers0