- Beginners - Haskell.org

Would you shed light on "undefined values" please?
by Costello, Roger L. 27 Jun '11

27 Jun '11

Hi Folks, In this book [1] the author defines the term "bottom": In order that we can say that, without exception, every syntactically well-formed expression denotes a value, it is convenient to introduce a special symbol (upside down T), pronounced 'bottom', to stand for the undefined value of a particular type. In particular, the value of infinity is the undefined value (bottom) of type Integer, and 1/0 is the undefined value (bottom) of type Float. Hence we can assert that 1/0 = bottom. He defines infinity as this: infinity :: Integer infinity = infinity + 1 The author says this when discussing the Bool datatype: It follows that there are not two but three Boolean values, namely False, True, and bottom. In fact, every datatype declaration introduces an extra anonymous value, the undefined value of the datatype. What is the undefined value (bottom) of type Bool? What is the undefined value (bottom) of type String? If I create my own datatype: data MyBool = F | T What is the undefined value (bottom) of type MyBool? I am not clear why "bottom" is an important concept. Would you explain please? /Roger [1] Introduction to Functional Programming using Haskell by Richard Bird

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Do I understand this well
by Roelof Wobben 27 Jun '11

27 Jun '11

Hello, Im stil reading chapter4 Now we have this : addVectors (x1, y1) (x2, y2) = (x1 + x2, y1 + y2) do I understand it that the outcome of x1+x2 will be saved in the last expression x1 + x2 But on which variable can the outcome be found. Roelof

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don't understand this
by Roelof Wobben 27 Jun '11

27 Jun '11

Hello, I must find the k st number in a list. So I have this list [1,2,3,4] and I have to find the second number. According to the solution the answer is : elementAt list i = list !! (i-1) But where do I fill in [1,2,3,4] and the 2 ? Roelof

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first number is a list
by Roelof Wobben 27 Jun '11

27 Jun '11

Hello, I have to find the first number is a list. So first I tried : let mylast = last [1,2,3,4]mylast This one works but you have to give 2 commands. So i was thinking about using list compreshion. So I did : [x | x <- [1,2,3,4] , last ] But then I see these error : Could not match expected type "Bool" with actual type [a0] -> a0 in the expression last in stdn of a list compreshion last in the expression : [x | x <- [1,2,3,4] , last ] What is my thinking error ? Roelof

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Re: [Haskell-beginners] [web-devel] yesod database select all rows?
by Michael Snoyman 27 Jun '11

27 Jun '11

On Mon, Jun 27, 2011 at 2:20 AM, Eric Schug <schugschug(a)gmail.com> wrote: > Hello, > It does not seem obvious to me how to perform the equivalent of > > SELECT * FROM table_name > SELECT field1,field2 FROM table_name > > examples show something of the form > rows <- runDB $ selectList [ Field1Eq value1, Field2Gt value2] [ ] 0 0 > But I don't see how the name of the table is determined nor how to select > additional fields to extract. > > What would this do? > rows <- runDB $ selectList [ ] [ ] 0 0 This all has to do with Haskell type trickery. If GHC can determine the expected type of rows based on how you use it, it will work. If it can't, it will complain about ambiguous types. If you want to select from the Person entity, you can do something like: rows <- runDB $ selectList ([] :: [Filter Person]) [] 0 0 Michael

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Re: [Haskell-beginners] Understanding some notation
by Jack Henahan 27 Jun '11

27 Jun '11

I know the type signature of (++), I'm just not seeing `n` as a list, I suppose. I'm reading it as being the Int that is passed to splitAt, though perhaps my thought process is just wonky. On Jun 26, 2011, at 6:41 PM, Mike Meyer wrote: > On Sun, 26 Jun 2011 18:14:44 -0400 > Jack Henahan <jhenahan(a)uvm.edu> wrote: > >> I have this code snippet: >> >> import Data.List >> >> aaa x (y:ys) = case splitAt x (y:ys) of >> (n, x:xs) -> x:n ++ xs >> (n, xs) -> n ++ xs >> >> I understand what it's meant to do (that is, split a list at index `x` and make a new list with that element at the head, or just return the list when given a singleton), but my brain is failing me when trying to read the notation `n ++ xs`. >> >> Is there some obvious explanation that I'm just forgetting? > > Yes. But more importantly, you're forgetting that you can ask the REPL: > > Prelude> :t (++) > (++) :: [a] -> [a] -> [a] > > So it takes two lists and produces a new one. Trying it: > > Prelude> [1, 2, 3] ++ [4, 5, 6] > [1,2,3,4,5,6] > > So ++ concatenates it's operands. > > <mike > -- > Mike Meyer <mwm(a)mired.org> http://www.mired.org/ > Independent Software developer/SCM consultant, email for more information. > > O< ascii ribbon campaign - stop html mail - www.asciiribbon.org ==== "Computer Science is no more about computers than astronomy is about telescopes." -- Edsger Dijkstra ====

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Understanding some notation
by Jack Henahan 27 Jun '11

27 Jun '11

I have this code snippet: import Data.List aaa x (y:ys) = case splitAt x (y:ys) of (n, x:xs) -> x:n ++ xs (n, xs) -> n ++ xs I understand what it's meant to do (that is, split a list at index `x` and make a new list with that element at the head, or just return the list when given a singleton), but my brain is failing me when trying to read the notation `n ++ xs`. Is there some obvious explanation that I'm just forgetting? ==== "Computer Science is no more about computers than astronomy is about telescopes." -- Edsger Dijkstra ====

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Need help understanding (1) typeclass instances that have multiple parameter types, and (2) overlapping instances
by Costello, Roger L. 26 Jun '11

26 Jun '11

Hi Folks, I have questions about: - typeclass instances that have multiple parameter types - overlapping instances Let's take an example of a typeclass with two instances: class MyShow a where toString :: a -> String instance MyShow Int where toString = show instance MyShow (Int, Int) where toString (a, b) = toString a ++ ", " ++ toString b The first instance has one parameter type (Int) and the second instance has two parameter types (Int, Int). When the class and instances are compiled, this error is generated: Illegal instance declaration for `MyShow (Int, Int)' (All instance types must be of the form (T a1 ... an) where a1 ... an are type *variables*, and each type variable appears at most once in the instance head. Use -XFlexibleInstances if you want to disable this.) In the instance declaration for `MyShow (Int, Int)' I interpret that error message to mean that Haskell has this rule: Instances with one parameter type can specify non-variable types (such as Int), but instances with multiple parameter types can only use variable types (e.g., a, b, c). Is that the rule? What is the rationale for that rule? Next, I placed this at the top of my file: {-# LANGUAGE FlexibleInstances #-} and the error message went away. Apparently the pragma is telling the compiler: Even though the Haskell code does not strictly conform to the Haskell specification, please let it compile. Is that what the pragma is telling the compiler? There must be a good reason why Haskell prohibits non-variable types in multi-parameter instances. So there must be a downside to adding that pragma. What is the downside? Now let's move on to my questions about overlapping instances. I created a third instance. It is for a pair of values of any type: instance (MyShow a, MyShow b) => MyShow (a, b) where toString (a, b) = ">>" ++ toString a ++ " " ++ toString b ++ "<<" It compiles without error. But if I try to use the second instance (the one that was defined for two Int's): toString ((34 :: Int), (44 :: Int)) then I get an overlapping instances error: Overlapping instances for MyShow (Int, Int) arising from a use of `toString' Matching instances: instance MyShow (Int, Int) -- Defined at Overlap.hs:11:9-25 instance (MyShow a, MyShow b) => MyShow (a, b) In the expression: toString ((34 :: Int), (44 :: Int)) I see how this: toString ((34 :: Int), (44 :: Int)) can match with the second and third instances, and thus the compiler has ambiguity on which instance to use. Is there a way to express, Hey compiler, please use this instance: instance MyShow (Int, Int) That is, when confronted with overlapping instances, what should I do? /Roger

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Fwd: Help
by Stoyan Peev 24 Jun '11

24 Jun '11

Yes, There are two way for me in this situation. The first one is to write function in another file for string list, that calls the parsing function, for every string in the list. But since now i have been trying in a couple of days and i can't do it successfully. The second options is just to rewrite this function for list of strings instead of just a string, but i couldn't do that either. So my question is can anybody help me finish it in one of the two ways. Thanks again! On Fri, Jun 24, 2011 at 11:11 PM, David Place <d(a)vidplace.com> wrote: > > On Jun 24, 2011, at 3:50 PM, Stoyan Peev wrote: > >> I'll be glad to finish the task myself, but i am going to need some help. > > Do you have a specific question? > > ____________________ > David Place > Owner, Panpipes Ho! LLC > http://panpipesho.com > d(a)vidplace.com > > > >

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Help
by Stoyan Peev 24 Jun '11

24 Jun '11

Hello all, I am experiencing some issues to do my course task in university. I have to write a calculator- function in Haskell. The function argument is a list of strings and also form such list, as each string of the argument made definite action: - If the string has the form of an arithmetic _expression_ - calculate this _expression_. The string result becomes part of the list-result. If the _expression_ contains a variable which is not assigned value, the result is displayed "undefined". - If the string has the form- Name = value calculated from the last _expression_ is assigned to the variable with the corresponding name in the list, and in the result list is formed a string with type - If there is not a calculated _expression_ to be assigned to form a string "no value". - If the string is non-blank, but there is a species different from the above two case, form the string "error". - If the string is empty, incl. when it contains only spaces, in the result there is not form a string. Expressions consist of integers without sign variables, operations + (Addition), - (subtraction), * (multiplication) and / (divide) and parentheses. Where no brackets, the operations are performed from left to right, but * and / precede the + and -. Implementation of any operation gives integer; in the division rejected the fractional part, if any. Variables have names of one letter - from the Latin small letter. In the beginning, end or between the elements of each row can have spaces - they are irrelevant to its correctness. Example: the list-argument ["3 +7 / 2" "2 + x", "= s", "2 * s +4", "", "2 + +4 / 5] function should provide a result-list ["6", "undefined", "s = 6", "16", "error"]. I say another person have the same task, but he didn't do anything. I started doing this task myself but i get stuck in the middle. Then i started searching for something that could help me and find out you :) The code i have written so far uses the library file "Parsing.lhs" but what i have written is taking those actions that i already described, only for a string. I cannot modify it to work for list of string, and complete the whole task. I'll be glad to finish the task myself, but i am going to need some help. Here is the code i have already written: import Parsing expr :: Parser Int expr = do t <- term do symbol "+" e <- expr return (t+e) +++ do symbol "-" e <- expr return (t-e) +++ return t term :: Parser Int term = do f <- factor do symbol "*" t <- term return (f * t) +++ do symbol "/" t <- term return (f-t) +++ return f factor :: Parser Int factor = do symbol "(" e <- expr symbol ")" return e +++ natural eval :: String -> Int eval xs = case (parse expr xs) of [(n,[])] -> n [(_,out)] -> error ("undefined") [] -> error "error" Thanks all in advance :) -- Best Wishes Stoyan Peev

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