Can you add or subtract two pointers?

[@kairu9373]

You deserve more subs!

[@martingeorgiev999]

8:47 the reason why -4 was printed is because anything allocated with malloc is aligned to a boundary that is suitable for all fundamental objects. Also, when you subtract pointers you get a value of type ptrdiff_t which is guaranteed to always be able to hold any pointer difference.

[@joseluisdominguezplaza6839]

Yeah I tried with an array of long long and had a different result (-1) that I think is the correct result

[@mavrix2356]

Amazing explanation 👌

[@novaz4768]

it wouldve been helpful to give us a work around. I was trying to add refrenced variables in a function

[@CodeVault]

For what do you need a workaround exactly? Can you explain more in detail so I understand what you need?

[@devDarshilJadav]

thank you so much

[@krtirtho]

I wish u were my professor

[@ayonsinha2075]

Sir u r are amazing ....💥🔥...where are from sir ??

[@CodeVault]

Thank you! I'm from Romania

[@m.a2894]

Hello, can you tell me why does my program returns '1' when I write it like this: int main() { long long int a = 15; long long int b = 20; long long int c = 30; long long res = &a - &c; printf(" %d\t%p", res, &res); } but returns '2' when i write it like this: int main() { long long int a = 15; long long int b = 20; long long int c = 30; printf("%p

[@CodeVault]

Huh, I am stumped as well. The only way to find out is to start looking at the differences in the assembly file. I'm sure there's something that happens there

[@nero2k619]

I guess its because &b has to pushed onto stack but when you try to comment out printf("%p

[@martingeorgiev999]

In the first case 'b' does not exist. The compiler is free to remove any junk from the code as long as it does not change the observable behavior.

[@mukteshgautam4050]

It adds 8bits not 8 bytes!

[@CodeVault]

The data type is a long long which is stored on 8 bytes. Adding any number to a pointer to long long would add that value times 8 bytes to the pointer. Remember, 1 byte = 8 bits. Not the other way around