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Comment by DannyBee | Hacker News Reader

No, it's correct. It's saying "if they are always computed, we move them up".

  if (a)
    foo = a + b
  else b 
    foo = a + b
  ->
  foo = a+b
  if (a)
  else (b)

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Possible side effects correction:

    ->
      if (a)
      else (b)
      foo = a+b

How do you move up towards exit?

I think you have to read it like "executed on >all paths to the function exit< as early as possible". As in, if you have an expression that is always executed regardless which path you take before returning. It's useful to move that expression upwards instead of having copies on the different paths to the function exit.

My guess is that they meant "up" in the sense of "up in post dominator tree".

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