Array manipulation (accessing specific bytes)

Question

I saw a video lecture about arrays and the lecturer said the output result of

int arr[5];
arr[3}=128;
((short *)arr)[6]=2;
cout << arr[3] << endl;

would be 512+128. I implement the code on my text editor, but after compile and run it the output is 2. My doubt is if my code is not correct or if the instructor is somehow wrong (or if I am misunderstanding some aspect).

My code:

#include <stdio.h>
int main(){
int array[5];
array[3]=128;
((short*)array)[6]=2;
printf("%d\n", array[3]);
return 0;
}

possible duplicate of When casting an int array to a short*, why does assigning a value to an element overwrite the entire integer? — Pankaj Andhale
– Pankaj Andhale, Commented Apr 7, 2015 at 19:26

Sergey Kalinichenko · Accepted Answer · 2015-04-07 20:00:07Z

3

the lecturer said the output result of [this code snippet] would be 512+128

The lecturer should have mentioned a few other things in order for the above to be true:

The result is dependent on hardware, which is big endian
The result is dependent on using a specific C compiler, with 16-bit ints and 8-bit shorts.

Otherwise, the result is not going to be the same.

My doubt is if my code is not correct or if the instructor is somehow wrong

The instructor is right, and your code is correct, but you are running this code on different platforms.

The main lesson that you should learn from this experiment is that code like that is fundamentally non-portable. If you wish to make it portable, use types from stdint.h to avoid sizing issues, and use bit manipulation instead of partial writes to avoid endianness problems.

edited Apr 7, 2015 at 20:00

answered Apr 7, 2015 at 19:25

Sergey Kalinichenko

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3 Comments

cooper Over a year ago

still not understanding why on my machine the output is 2. If I'm changing the value of "short 6" , doesn't that imply that array[3] remains with value 128? I mean, "short 6" is inside array[2] , right? Can you help me? How do I know my machine's endieness?

Sergey Kalinichenko Over a year ago

@cooper You probably have a 32-bit, little endian platform with 16-bit shorts. Because of this, int values that are smaller than 16 bits would be the same when you retrieve them from the same address as 32 or 16 bits. The best exercise to do would be drawing your int array as 20 adjacent squares (4 bytes per int), writing the value 128 into square 12 (zero-based; bytes 13, 14, and 15 are zeros) and then looking at the content of bytes 12 and 13 - the two bytes of short at index 6. They would have 128 before the replacement, and 2 after the replacement.

Sergey Kalinichenko Over a year ago

@cooper Another useful exercise is to replace values with something that occupies the entire 32/16 bits, and see what's going to happen (demo).

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