How to implement a hash table in C (2021) • Hacker News Board

Re: How to implement a hash table in C (2021)

#41

Post by rerdavies » Mon, Jul 22, 2024, 11:31 PM UTC

Joker_vD wrote: post #40

rerdavies wrote: post #39

Earlier quoted context omitted.

Interesting. I hadn't thought of that. ... although I can't help thinking you have larger problems if you have two objects larger than 16k in the same array in a machine that has at most 64k of memory. :-P Or a char array greater than 32k for that matter. I do appreciate the theoretical horror, but thank you anyway, C23! Having to cast your size_t's to ptrdiff_t's before you subtract them in order to preserve portabi…

You always could subtract two size_t's: they're unsigned, after all, and such behaviour always have been defined, and the result is also unsigned. But if you want to subtract pointers, apparently you better convert them to uintptr_t first.

lol. You can't always subtract two size_t's. (size_t)3 - (size_t)4. And (ptrdiff_t)((size_t)3-(size_t)4) actually would be undefined behavior in C++20!

Re: How to implement a hash table in C (2021)

#42

Post by Joker_vD » Wed, Jul 24, 2024, 2:03 PM UTC

rerdavies wrote: post #41

Joker_vD wrote: post #40

Earlier quoted context omitted.

You always could subtract two size_t's: they're unsigned, after all, and such behaviour always have been defined, and the result is also unsigned. But if you want to subtract pointers, apparently you better convert them to uintptr_t first.

lol. You can't always subtract two size_t's. (size_t)3 - (size_t)4. And (ptrdiff_t)((size_t)3-(size_t)4) actually would be undefined behavior in C++20!

The "(size_t)3 - (size_t)4" is defined to be the same as "-(size_t)1", which is defined, and the same as "(size_t)1 + ~(size_t)1". Also, "(ptrdiff_t)((size_t)3-(size_t)4)" is defined both before and after C++20:

    If the destination type is signed, the value does not change if the source integer can be represented in the destination type. Otherwise the result is [implementation-defined (until C++20)] [the unique value of the destination type equal to the source value modulo 2^n where n is the number of bits used to represent the destination type (since C++20)] (note that this is different from signed integer arithmetic overflow, which is undefined).

Re: How to implement a hash table in C (2021)

#43

Post by rerdavies » Tue, Jul 30, 2024, 9:02 AM UTC

Joker_vD wrote: post #42

rerdavies wrote: post #41

Earlier quoted context omitted.

lol. You can't always subtract two size_t's. (size_t)3 - (size_t)4. And (ptrdiff_t)((size_t)3-(size_t)4) actually would be undefined behavior in C++20!

The "(size_t)3 - (size_t)4" is defined to be the same as "-(size_t)1", which is defined, and the same as "(size_t)1 + ~(size_t)1". Also, "(ptrdiff_t)((size_t)3-(size_t)4)" is defined both before and after C++20: If the destination type is signed, the value does not change if the source integer can be represented in the destination type. Otherwise the result is [implementation-defined (until C++20)] [the unique value…

However, the value of (ptrdiff_t)((size_t)3-(size_t)4) is positive if ptrdiff_t has at least 17 bits, and size_t only has 16 bits, and negative whenever ptrdiff_t and size_t have the same number of bits.