5 ms·

Isn't this kind of negated by the fact that `delete arr[n]` working the way it does has required JavaScript arrays to never have been "real" arrays, in the enti

Isn't this kind of negated by the fact that `delete arr[n]` working the way it does has required JavaScript arrays to never have been "real" arrays, in the entire history of the language?

eyelidlessness 2y ago
I don’t think so. There’s a little bit more nuance to it (because it involves array “holes”—almost a third kind of null), but delete on an array index is semantically closer to assigning undefined to the index. It doesn’t actually delete the presence of the index.
- Karellen 2y ago
  > It doesn’t actually delete the presence of the index. Doesn't it? Try: a = [ 6, 7, 8, 9, 10 ] delete a[2] for (x in a) { console.log(x) }
  - eyelidlessness 2y ago
    It doesn’t. It deletes the property “2”, but the index 2 remains present as a “hole” (like I mentioned before, almost a third kind of null). That removes it from iteration over the array object’s properties (i.e. your for-in loop), but doesn’t remove the index from the array or its iterable value representation: https://www.typescriptlang.org/play/?#code/MYewdgziA2CmB0w4EMBOAKAlAWAFB9EgBcACZEgXhIG0SA2AGhIHYmAOJgTiYEYAGEgF0A3HgLgoceNBABzdACIARrABmIVLAVNk02GFlEAFjpz5c61CXSEIpAB4kQqsphIBvPCW8lbMBDLy9jrwAJZgACaw9gDyquj2mGYAvmLmUXBEsGTUAEyCaX5SgYrIqlmo2mR6BsamaZbWtg5OLshunrg+vhL+0nIJIeFRsfGJKXhAA https://www.typescriptlang.org/play/?#code/MYewdgziA2CmB0w4E... If you click run there and look at the logged output, you’ll see: 1. The length remains the same after the delete statement 2. The items after index 2 keep their index 3. The for-of loop continues to iterate over index 2 (which now produces an undefined value on read)
    - Karellen 2y ago
      > the index 2 remains present as a “hole” (like I mentioned before, almost a third kind of null). It's an interesting model, and I think it definitely has its uses when reasoning about JS arrays. I'm not yet convinced it's the best model, but it's something I'm going to have to let percolate through my brain for a while to weigh it up properly. Thanks.
    - Dylan16807 2y ago
      I think you're misinterpreting how for-of works. For-of will iterate over deleted keys in the exact same way that it iterates over keys that never existed. let a = []; a[7] = 7; for (let v of a) console.log(v); > undefined, undefined, undefined, undefined, undefined, undefined, undefined, 7
- recursive 2y ago
  > It doesn’t actually delete the presence of the index. It actually removes the index. > Object.keys([9,,9]) [ "0", "2" ] The value of the `length` property does nothing to imply the presence or lack of any keys. This is covered in the "exotic objects" behavior section in chapter 10 of the ECMAscript spec.
  - eyelidlessness 2y ago
    It removes the property key. It does not remove the index. You can see the distinction in the iterator produced by the array prototype’s keys method which, per ECMAScript spec chapter 23, yields the array’s indexes. And they are, in fact, determined by the array’s length. You can delete every property corresponding to an array’s indexes, and you’ll get the same result.

Isn't this kind of negated by the fact that `delete arr[n]` working the way it does has required JavaScript arrays to never have been "real" arrays, in the entire history of the language?

eyelidlessness 2y ago

I don’t think so. There’s a little bit more nuance to it (because it involves array “holes”—almost a third kind of null), but delete on an array index is semantically closer to assigning undefined to the index. It doesn’t actually delete the presence of the index.

Karellen 2y ago
> It doesn’t actually delete the presence of the index. Doesn't it? Try: a = [ 6, 7, 8, 9, 10 ] delete a[2] for (x in a) { console.log(x) }
- eyelidlessness 2y ago
  It doesn’t. It deletes the property “2”, but the index 2 remains present as a “hole” (like I mentioned before, almost a third kind of null). That removes it from iteration over the array object’s properties (i.e. your for-in loop), but doesn’t remove the index from the array or its iterable value representation: https://www.typescriptlang.org/play/?#code/MYewdgziA2CmB0w4EMBOAKAlAWAFB9EgBcACZEgXhIG0SA2AGhIHYmAOJgTiYEYAGEgF0A3HgLgoceNBABzdACIARrABmIVLAVNk02GFlEAFjpz5c61CXSEIpAB4kQqsphIBvPCW8lbMBDLy9jrwAJZgACaw9gDyquj2mGYAvmLmUXBEsGTUAEyCaX5SgYrIqlmo2mR6BsamaZbWtg5OLshunrg+vhL+0nIJIeFRsfGJKXhAA https://www.typescriptlang.org/play/?#code/MYewdgziA2CmB0w4E... If you click run there and look at the logged output, you’ll see: 1. The length remains the same after the delete statement 2. The items after index 2 keep their index 3. The for-of loop continues to iterate over index 2 (which now produces an undefined value on read)
  - Karellen 2y ago
    > the index 2 remains present as a “hole” (like I mentioned before, almost a third kind of null). It's an interesting model, and I think it definitely has its uses when reasoning about JS arrays. I'm not yet convinced it's the best model, but it's something I'm going to have to let percolate through my brain for a while to weigh it up properly. Thanks.
  - Dylan16807 2y ago
    I think you're misinterpreting how for-of works. For-of will iterate over deleted keys in the exact same way that it iterates over keys that never existed. let a = []; a[7] = 7; for (let v of a) console.log(v); > undefined, undefined, undefined, undefined, undefined, undefined, undefined, 7
recursive 2y ago
> It doesn’t actually delete the presence of the index. It actually removes the index. > Object.keys([9,,9]) [ "0", "2" ] The value of the `length` property does nothing to imply the presence or lack of any keys. This is covered in the "exotic objects" behavior section in chapter 10 of the ECMAscript spec.
- eyelidlessness 2y ago
  It removes the property key. It does not remove the index. You can see the distinction in the iterator produced by the array prototype’s keys method which, per ECMAScript spec chapter 23, yields the array’s indexes. And they are, in fact, determined by the array’s length. You can delete every property corresponding to an array’s indexes, and you’ll get the same result.