Linus Torvalds' good taste argument for linked lists, explained

Post by lmm » Mon, Dec 07, 2020, 3:24 AM UTC

macspoofing wrote: post #7

I understand the general point (and value) of reframing the problem or the solution in a way that removes special cases ... but in this case I would actually prefer the first solution over the second. The second solutions reminds me of the old-school perl culture, and JavaScript culture, where 'cleverness' (which always manifests itself as terseness as if lines of code were expensive), takes precedence over maintaina…

Post by btilly » Mon, Dec 07, 2020, 3:38 AM UTC

andy_threos_io wrote: post #155

jariel wrote: post #91

Earlier quoted context omitted.

You make a very good point, and that it has to do with the relative foundational conceptualisational ability of the types of mainteners. That's great because it kind of speaks to the crux of the problem. But the first solution does use pointers :) The second uses double pointers. I'd argue that 'even kernel maintainers' may not be so easy with the second in reality. It's probably worth it if there is a performance ga…

The simple solution is faster.

Post by a1369209993 » Mon, Dec 07, 2020, 3:46 AM UTC

alisonkisk wrote: post #77

R0b0t1 wrote: post #57

Earlier quoted context omitted.

I was only ever taught the second, more concise one, so this discussion is kind of confusing. How is it not shorter and more explicit?

Many Programmers who don't use C are afraid of pointers and prefer superficially simpler (but overall more complicated) things like Java's abstraction towers, despite the extra expense.

Post by zgs » Mon, Dec 07, 2020, 3:58 AM UTC

Post by cambalache » Mon, Dec 07, 2020, 4:13 AM UTC

brainwipe wrote: post #33

nemetroid wrote: post #19

I'm not sure why the article removed comments from the code and replaced variable names like "indirect" with "p". Here are the two code samples verbatim from Linus's presentation: remove_list_entry(entry) { prev = NULL; walk = head; // Walk the list while (walk != entry) { prev = walk; walk = walk->next; } // Remove the entry by updating the // head or the previous entry if (!prev) head = entry->next; else prev->next…

Thank you, I found that much easier to read than the article posted.