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In the computer, you also have no descriptive variable names and make excessive use of gotos. This is far from a good argument. > Haskell does support mutatio

In the computer, you also have no descriptive variable names and make excessive use of gotos. This is far from a good argument. > Haskell does support mutation, but inside of particular monads. You are mistaken. Monads do not involve mutation; that would defeat their entire purpose. Although it's true that the monadic code is being compiled into assembly that involves mutation, just as $LANGUAGE code is being compiled into assembly that includes a lot of goto.

tome 12y ago
> You are mistaken. Monads do not involve mutation. Indeed. IO and ST involve mutation. They just so happen to be instances of Monad, but that's tangential.
- rtpg 12y ago
  they don't involve mutation on the language semantics level. The mutation happens through the runtime (via calls to the C FFI). In reality they're implemented in a mutation-heavy way for performaance reasons, but on the language level absolutely no mutation happens ( it's the whole World -> (Result, World) analogy)
  - carterschonwald 12y ago
    They're not done via cffi . its all compiler supported primops that get optimized and lowered to target native assembly.
  - theseoafs 12y ago
    The whole point about where the mutation happens and how the language remains pure in spite of it is not awfully relevant. The point is that Haskell does support real, bona-fide mutation in its standard library (see for example its support for stateful IO arrays). So obviously mutation is useful, regardless of whether you're doing that mutation behind a pure interface.
- BillMryes 12y ago
  To add to what rtpg says, if you look at an actual, monadic example of an IO: putStrLn "What is your name?" >> getLine >>= \name -> putStrLn ("Hi, " ++ name ++ "!") It's just a giant chain of instructions, each previous result feeding into the next. Nowhere do we assign anything, modify anything, etc. No mutation.
tel 12y ago
This is said a lot and technically valid in a sense, but I think over-pedantic. Some monads, IO, ST, State, do support mutation in the internal language. Then their implementation, their representation in the meta language, Haskell, can be "pure or whatever who cares". http://www.reddit.com/r/haskell/comments/30l46z/haskell_for_all_algebraic_side_effects/cptlduv
im2w1l 12y ago
It's an implementation of mutability in a non-mutable language implemented on a platform with mutability.

tome 12y ago

> You are mistaken. Monads do not involve mutation. Indeed. IO and ST involve mutation. They just so happen to be instances of Monad, but that's tangential.

rtpg 12y ago
they don't involve mutation on the language semantics level. The mutation happens through the runtime (via calls to the C FFI). In reality they're implemented in a mutation-heavy way for performaance reasons, but on the language level absolutely no mutation happens ( it's the whole World -> (Result, World) analogy)
- carterschonwald 12y ago
  They're not done via cffi . its all compiler supported primops that get optimized and lowered to target native assembly.
- theseoafs 12y ago
  The whole point about where the mutation happens and how the language remains pure in spite of it is not awfully relevant. The point is that Haskell does support real, bona-fide mutation in its standard library (see for example its support for stateful IO arrays). So obviously mutation is useful, regardless of whether you're doing that mutation behind a pure interface.
BillMryes 12y ago
To add to what rtpg says, if you look at an actual, monadic example of an IO: putStrLn "What is your name?" >> getLine >>= \name -> putStrLn ("Hi, " ++ name ++ "!") It's just a giant chain of instructions, each previous result feeding into the next. Nowhere do we assign anything, modify anything, etc. No mutation.

tel 12y ago

This is said a lot and technically valid in a sense, but I think over-pedantic. Some monads, IO, ST, State, do support mutation in the internal language. Then their implementation, their representation in the meta language, Haskell, can be "pure or whatever who cares". http://www.reddit.com/r/haskell/comments/30l46z/haskell_for_all_algebraic_side_effects/cptlduv

im2w1l 12y ago

It's an implementation of mutability in a non-mutable language implemented on a platform with mutability.