5 ms·

> Your (1) only affects compilation time, [...] No, it mostly affects execution time. Whether the effect is meaningful, however, depends on various factors. >

by adwn 6y ago

> Your (1) only affects compilation time, [...] No, it mostly affects execution time. Whether the effect is meaningful, however, depends on various factors. > [...] and even this is assuming a particular implementation of arrays (which is not a given). Eventually, all arrays boil down to pointer+offset memory accesses. Well, if your programming language internally implements "arrays" as hashmaps, then you've already lost performance-wise, of course. > Why would you need to use "modulo and division into round/ceil/floor" on array offsets? Whenever you're working with 2D/3D data in manually managed memory. > How often you do it? When you're writing web services? Probably not so much. When you're doing image manipulation or 3D graphics? All the time.

xxs 6y ago
>When you're writing web services? Probably not so much. I bet it's done a lot as a lot of data structures are implemented vid pow2 length arrays. It's just the people who write them are unaware.
3JPLW 6y ago
> No, it mostly affects execution time. Citation needed. Yes, the -1 appears in the source code. It might even appear in the intermediate representation. It'll even show up in the assembly for multidimensional arrays (but it'll all get folded into a single `mov` assembly for vectors). But I dare you to actually demonstrate a performance difference. Compared to a memory access, a `dec` is free.
nvader 6y ago
At some point I would like to write up a story about this team in my University Object-Oriented Programming class who decided to use 1-based 1ndexes in Java while making a 2-D array for a board game. I don't want to spoil it, but the following picture of a how they stored a 3x3 tic-tac-toe board is a hint: [ null, null, null, null, null, null, null, X, 0, X, null, null, null, null, O, X, X, null, null, null, null, X, 0, 0 ]
professoretc 6y ago
> No, it mostly affects execution time. That's what I was thinking, too, but most of the time (even if the index is not static) the -1 can be eliminated by just modifying the address at which the array is assumed to start. That is, arr[x] compiles into an access at address (arr - 1) + x Since the address of arr is statically known, arr - 1 is, too, and the address of "one before the first element" will be what's written as the base for the memory access.
- adwn 6y ago
  > Since the address of arr is statically known [...] No, that's almost never the case – only when accessing a static array. Typically, that restricts it to static look-up tables and very simple embedded systems.

by adwn 6y ago

xxs 6y ago

>When you're writing web services? Probably not so much. I bet it's done a lot as a lot of data structures are implemented vid pow2 length arrays. It's just the people who write them are unaware.

3JPLW 6y ago

> No, it mostly affects execution time. Citation needed. Yes, the -1 appears in the source code. It might even appear in the intermediate representation. It'll even show up in the assembly for multidimensional arrays (but it'll all get folded into a single `mov` assembly for vectors). But I dare you to actually demonstrate a performance difference. Compared to a memory access, a `dec` is free.

nvader 6y ago

At some point I would like to write up a story about this team in my University Object-Oriented Programming class who decided to use 1-based 1ndexes in Java while making a 2-D array for a board game. I don't want to spoil it, but the following picture of a how they stored a 3x3 tic-tac-toe board is a hint: [ null, null, null, null, null, null, null, X, 0, X, null, null, null, null, O, X, X, null, null, null, null, X, 0, 0 ]

professoretc 6y ago

> No, it mostly affects execution time. That's what I was thinking, too, but most of the time (even if the index is not static) the -1 can be eliminated by just modifying the address at which the array is assumed to start. That is, arr[x] compiles into an access at address (arr - 1) + x Since the address of arr is statically known, arr - 1 is, too, and the address of "one before the first element" will be what's written as the base for the memory access.

adwn 6y ago
> Since the address of arr is statically known [...] No, that's almost never the case – only when accessing a static array. Typically, that restricts it to static look-up tables and very simple embedded systems.