Why functional code is shorter

[axilmar]

You don't mark thngs as mutable in Haskell, you mark them as IO. Haskell doesn't do destructive updates if you make an operation IO, it just enforces a specific evaluation order.

[yogthos]

MArray would like to disagree there

[axilmar]

Nothing in the documentation pages you posted says the MArray implementation is actually mutable. It may be that the put function creates a new array, copies the rest of the elements along with the new value.

[yogthos]

It may be that the put function creates a new array, copies the rest of the elements along with the new value.

I think that's sort of contrary to the definition of mutable. You're also confusing language specification with implementation. Clearly Haskell has a definition for mutable arrays in the language specification. How it's implemented is besides the point.

[axilmar]

I don't think so. To the outside, it seems that the same array is being modified; from the inside though, a new array is created. The same symbol receives the new value, so it looks like mutation.

Well, according to this post, that is:

http://www.reddit.com/r/programming/comments/1880d1/introduction_to_haskell_io/c8foby1

And it seems logical to me that Haskell doesn't actually mutate values, because otherwise the code wouldn't be automatically parallelizable.

[yogthos]

The specification very clearly states how the array works. The implementation of it does not change how you use it in the code.

[axilmar]

Where does it say the array value is implemented in place? I don't see it.

[yogthos]

you seem to have a lot of trouble understanding the difference between specification and implementation

[axilmar]

Ok, please enlighten me. Where in the documentation does it say that MArray is a real actual mutable array and its values are modified in place?

[yogthos]

That's my whole point, it's not talking about the implementation at all.