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haskellghcinlining

Do INLINE pragmas cause inlining even if optimizations are disabled?

GHC docs on INLINE pragmas:

GHC (with -O, as always) tries to inline (or “unfold”) functions/values that are “small enough,” [...]

The major effect of an INLINE pragma is to declare a function’s “cost” to be very low. The normal unfolding machinery will then be very keen to inline it. [...]

I take this to mean that without optimizations enabled, even {-# INLINE ... #-} functions / values will not be inlined, since the "normal unfolding machinery" will not even be run to notice the function's cost has been reduced.

Is that correct?

Solution

  • The reason this happens is that -O0 still invokes a single pass of the "simplifier", which is the optimization pass responsible for, among other things, inlining.

    You can't prevent the simplifier from being called, but you can set the number of simplifier iterations to zero, which should cause it to bail out before doing anything:

    ghc -O0 -fmax-simplifier-iterations=0 -ddump-simpl

    For your handleThing example, this prevents the inlining you've observed.

    The inlining that occurs with -O0 is limited to what can be accomplished by inlining fully saturated calls with no eta expansion/reduction. This means that your example:

    handleThing = print
{-# INLINE handleThing #-}
main = mapM_ handleThing [1..10]

    will inline, but the variation:

    handleThing x = print x
{-# INLINE handleThing #-}
main = mapM_ handleThing [1..10]

    won't, since the call to handleThing in main is no longer fully saturated.

    On the other hand, this will inline both number and handleThing:

    number = 10
{-# INLINE number #-}
handleThing x = print x >> print x
{-# INLINE handleThing #-}
main = handleThing number

    into main, resulting in:

    x = 10
main = >> $fMonadIO (print $fShowInteger x) (print $fShowInteger x)

    Here, "inlining" number involves the de-optimization of giving it an extra name and using that.