Addition of unsigned long and int in C

Question

    int meta_size = 24;
    node_t* prev;

    printf("%lx, ", prev + meta_size );
    printf("%lx, ", prev); 
    printf("%lx, ", meta_size);

out put: 1519240, 1519000, 18 how this is happening?

What did you expect? You are using the wrong format specifier to print a pointer, which invokes undefined baheviour. And despite your title, you don't add a `unsigned long` and an `int`. — too honest for this site, Apr 16 '16 at 19:31
Do you understand [hexadecimal](https://en.wikipedia.org/wiki/Hexadecimal) numbers? — user3386109, Apr 16 '16 at 19:33

score 2 · Accepted Answer · answered Apr 16 '16 at 19:34

Prev is a unitialised pointer, it contains (a random, because it is not initialised) memory address.

printf("%lx, ", prev + meta_size );//Prints the memory address prev is pointing to + (sizeof(node_t) * meta_size)
printf("%lx, ", prev);             //Prints the memory address prev is pointing to
printf("%lx, ", meta_size);        //Prints meta_size, 18 is 24 in hexidecimal, because of the 'x' in %lx

However, the first 2 lines are undefined behaviour because pointers should be printed with %p

prev is initialized. I just didn't show it. – canbax Apr 17 '16 at 03:11 — canbax, Apr 17 '16 at 03:11

score 0 · Answer 2 · answered Apr 16 '16 at 19:33

0

if you do sizeof(node_t) I'm pretty sure you will obtain 10.

answered Apr 16 '16 at 19:33

Dinsdale

184
3
12

1

Actually `24` decimal or `0x18` hex. Note that all of the output numbers are hex. – user3386109 Apr 16 '16 at 19:38
yes you are right. When you add n to a node_t pointer, the pointer address will be increase by n * (sizeof(node_t)) thats why it is increase by 240 in base 16 = 576 in base 10 = 24 * 24 in base 10. – canbax Apr 17 '16 at 03:09

Addition of unsigned long and int in C

2 Answers2