Bit Size of GHC's Int Type

问题:

Why is GHC's Int type not guaranteed to use exactly 32 bits of precision? This document claim it has at least 30-bit signed precision. Is it somehow related to fitting Maybe Int or similar into 32-bits?

回答1:

It is to allow implementations of Haskell that use tagging. When using tagging you need a few bits as tags (at least one, two is better). I'm not sure there currently are any such implementations, but I seem to remember Yale Haskell used it.

Tagging can somewhat avoid the disadvantages of boxing, since you no longer have to box everything; instead the tag bit will tell you if it's evaluated etc.

回答2:

The Haskell language definition states that the type Int covers at least the range [−2²⁹, 2²⁹−1]. There are other compilers/interpreters that use this property to boost the execution time of the resulting program.

All internal references to (aligned) Haskell data point to memory addresses that are multiple of 4(8) on 32-bit(64-bit) systems. So, references need only 30bits(61bits) and therefore allow 2(3) bits for "pointer tagging".

In case of data, the GHC uses those tags to store information about that referenced data, i.e. whether that value is already evaluated and if so which constructor it has.

In case of 30-bit Ints (so, not GHC), you could use one bit to decide if it is either a pointer to an unevaluated Int or that Int itself.

Pointer tagging could be used for one-bit reference counting, which can speed up the garbage collection process. That can be useful in cases where a direct one-to-one producer-consumer relationship was created at runtime: It would result directly in memory reuse instead of a garbage collector feeding.

So, using 2 bits for pointer tagging, there could be some wild combination of intense optimisation... In case of Ints I could imagine these 4 tags:

• a singular reference to an unevaluated Int
• one of many references to the same possibly still unevaluated Int
• 30 bits of that Int itself
• a reference (of possibly many references) to an evaluated 32-bit Int.

回答3:

I think this is because of early ways to implement GC and all that stuff. If you have 32 bits available and you only need 30, you could use those two spare bits to implement interesting things, for instance using a zero in the least significant bit to denote a value and a one for a pointer.

Today the implementations don't use those bits so an Int has at least 32 bits on GHC. (That's not entirely true. IIRC one can set some flags to have 30 or 31 bit Ints)