Comment by edflsafoiewq

13 hours ago

Doesn't reduce force the accumulator to be shared though? Both the reduce and the lambda are holding onto references to acc, which defeats any "single reference" optimizations.

8 comments

edflsafoiewq

vhcr 9 hours ago

The problem with:

    ret = ""
    for s in strings:
        ret += s

is that it re-allocates O(n) times, even if ret is referenced only once.

edflsafoiewq 9 hours ago

If the s are small the usual geometric buffer growth mitigates that. Of course you can compute the final buffer size in this case, but often you have a bunch of dynamically-generated strings of different sizes.

ndriscoll 12 hours ago

  def reduce(acc, f): 
    for v in self:
      acc = f(acc, v)
    return acc

The current acc goes out of scope each time you call f. There's no shared reference (assuming f doesn't sneak store it elsewhere, which for string combining, f should just be `return a+b`?).

edflsafoiewq 11 hours ago
The binding for acc in the reduce call is still active during the f call, which means there are at least two references to acc.
- ndriscoll 11 hours ago
  
  Why is it still active? Even an interpreter with no lookahead could see that it goes out of scope immediately when f returns (it gets shadowed on that line), so as long as there's no guarantee about when finalizers get called, it should be able to mark it dead inside of reduce as soon as it's passed to f. Like move semantics here should be a general pattern for optimization, no?
  
  2 replies →

Zak 12 hours ago

It might - let's assume it does. My point is that it's better to use the explicit optimized method for joining strings in a performance-sensitive context than to try to meet the conditions for an implicit optimization.