On Tuesday, 2003-08-12, 12:14, CEST, Alistair Bayley wrote:
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Care to explain why you think Haskell is not pure?
Purely functional = you are always defining and calling pure functions. Calling a pure function with the same arguments always gives the same results. In Haskell, if monadic state is used, then this is not true (there is support from the syntax + type system to hide the state argument). So Haskell's monadic state is an implementation of state *on top of* a purely functional language.
Hello, I wouldn't agree. In my opinion, in Haskell a function always gives the same result when called with the same arguments, even if it involves I/O. So Haskell is truely purely functional. The point is that evaluating an expression of an IO type doesn't actually do the I/O and yield the result of the I/O action. It yields a result which *describes the action* to do. The I/O is only finally done because the meaning of a Haskell program is to execute the action described by main. For example, the function readFile is pure. For a specific string s the expression readFile s always yields the same result: an I/O action which searches for a file named s, reads its content and takes this content as the result of the *action* (not the expression). Correct me if I'm wrong. I wouldn't say that I'm an expert in these things.
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Wolfgang