Roy Feague

Roy Feague rfeague

Niner since 2009


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Side-by-side screwup



  • C9 Lectures: Dr. Erik Meijer - Functional Programming Fundamentals Chapter 4 of 13

    Thanks ShinNoNoir and exoteric.  

    This is analogous to correlated subqueries in SQL, which also tax my brain.


    I guess what I find weird about \x -> (\y -> x+y) is:

    1. the syntax suggests to me that the definition of \y is complete within the parentheses, but it has a dependency that reaches outside the parens and raises questions about the limits of scope; and (related)

    2. in other programming contexts one learns to read parentheses from innermost to outermost.  But here we need to read it left to right.


    I guess it's just going to take a while to learn to think like the Haskell parser. 


    Another example that twists my C/C++ programmer's brain: we can define tail as

       mytail (_:xs) = xs

    Very logical, but this assumes that the language is able to deconstruct the list input to understand that it can be thought of as a single element added to a list.  Cool that the language can do that, but not obvious, and leaves me wondering what other sorts of deconstructions it is capable of making that are not obvious.  Can I do drop2 (b:(a:xs)) = b:xs to drop the second item?  How will we know the limits of such constructs without just trial and error?  I'm sure this will become more clear as we go on.


    re: (-1) 3 --- agreed that the problem is that -1 can be read (and apparently IS read) as negative 1.  Surprising then that we can't use ((-)1) 3 or something analogous.  The flip business seems very convoluted.  But I suppose this is a minor syntactic quirk originating from the overloaded use of the - symbol, and since it only impacts a syntax shortcut, it's probably not worth worrying too much about.


  • C9 Lectures: Dr. Erik Meijer - Functional Programming Fundamentals Chapter 4 of 13

    I am delighted to have found this series. Really great stuff, and it's a real pleasure to have access to such a top-tier instructor.  Thanks Erik!


    In the function

        add = \x -> (\y -> x+y)

    how can the inner function \y access the value of x when it is defined to take only a single parameter y?


    And why does the expression

       (-1) 3

    return an error, but

      (/2) 6

    is fine?