instance for (Show ([(String, Int)] -> Int))

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How to "instance MonadIO Identity"?

Aaron Gray

24 Dec 2010 24 Dec '10

1:17 p.m.

How do I code an 'instance' declaration for '(Show ([(String, Int)] -> Int))'. I am new to instance declarations and am following the Happy examples :- http://www.haskell.org/happy/doc/html/sec-using.html#sec-other-datatypes The compiler is requesting an instance declaration for Show :- expr-eval.hs:334:23: No instance for (Show ([(String, Int)] -> Int)) arising from a use of `print' at expr-eval.hs:334:23-27 Possible fix: add an instance declaration for (Show ([(String, Int)] -> Int)) In the first argument of `(.)', namely `print' In the second argument of `(>>=)', namely `print . calc . lexer' In the expression: getContents >>= print . calc . lexer Many thanks in advance, Aaron

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Henning Thielemann

24 Dec 24 Dec

1:23 p.m.

On Fri, 24 Dec 2010, Aaron Gray wrote:

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How do I code an 'instance' declaration for '(Show ([(String, Int)] -> Int))'.

You ask for displaying a function that maps from [(String, Int)] to Int. What shall the generated text look like?

Henning Thielemann

1:24 p.m.

On Fri, 24 Dec 2010, Aaron Gray wrote:

...

The compiler is requesting an instance declaration for Show :-

expr-eval.hs:334:23: No instance for (Show ([(String, Int)] -> Int)) arising from a use of `print' at expr-eval.hs:334:23-27 Possible fix: add an instance declaration for (Show ([(String, Int)] -> Int)) In the first argument of `(.)', namely `print' In the second argument of `(>>=)', namely `print . calc . lexer' In the expression: getContents >>= print . calc . lexer

... maybe 'calc' needs a further argument?

Aaron Gray

4:47 p.m.

On 24 December 2010 18:24, Henning Thielemann

...

wrote:

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On Fri, 24 Dec 2010, Aaron Gray wrote:

The compiler is requesting an instance declaration for Show :-

...
expr-eval.hs:334:23: No instance for (Show ([(String, Int)] -> Int)) arising from a use of `print' at expr-eval.hs:334:23-27 Possible fix: add an instance declaration for (Show ([(String, Int)] -> Int)) In the first argument of `(.)', namely `print' In the second argument of `(>>=)', namely `print . calc . lexer' In the expression: getContents >>= print . calc . lexer

... maybe 'calc' needs a further argument?

I have attached what I have typed in so far. Aaron

Daniel Fischer

5:07 p.m.

On Friday 24 December 2010 22:47:55, Aaron Gray wrote:

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On 24 December 2010 18:24, Henning Thielemann
...
wrote:

On Fri, 24 Dec 2010, Aaron Gray wrote:

The compiler is requesting an instance declaration for Show :-

...
expr-eval.hs:334:23: No instance for (Show ([(String, Int)] -> Int)) arising from a use of `print' at expr-eval.hs:334:23-27 Possible fix: add an instance declaration for (Show ([(String, Int)] -> Int)) In the first argument of `(.)', namely `print' In the second argument of `(>>=)', namely `print . calc . lexer' In the expression: getContents >>= print . calc . lexer

... maybe 'calc' needs a further argument?

I have attached what I have typed in so far.

Well, *ExprEval> :t calc calc :: [Token] -> [(String, Int)] -> Int calc needs an environment (a dictionary of let-bound names), which you have to provide. main = print . flip calc [] . lexer works fine.

...

Aaron

Aaron Gray

6:32 p.m.

On 24 December 2010 22:07, Daniel Fischer wrote:

...

On Friday 24 December 2010 22:47:55, Aaron Gray wrote:

...
On 24 December 2010 18:24, Henning Thielemann
...
wrote:

On Fri, 24 Dec 2010, Aaron Gray wrote:

The compiler is requesting an instance declaration for Show :-

...
expr-eval.hs:334:23: No instance for (Show ([(String, Int)] -> Int)) arising from a use of `print' at expr-eval.hs:334:23-27 Possible fix: add an instance declaration for (Show ([(String, Int)] -> Int)) In the first argument of `(.)', namely `print' In the second argument of `(>>=)', namely `print . calc . lexer' In the expression: getContents >>= print . calc . lexer

... maybe 'calc' needs a further argument?

I have attached what I have typed in so far.

Well,

*ExprEval> :t calc calc :: [Token] -> [(String, Int)] -> Int

calc needs an environment (a dictionary of let-bound names), which you have to provide.

main = print . flip calc [] . lexer

works fine.

Okay great, works this end too, but what does the 'flip' do ??? Now to get it to work as a REPL and to read from file :) Aaron

Daniel Fischer

6:58 p.m.

On Saturday 25 December 2010 00:32:38, Aaron Gray wrote:

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Okay great, works this end too, but what does the 'flip' do ???

It flips the order of arguments to calc. You could also write main = getContents >>= print . (`calc` []) . lexer Generally, flip f = \x y -> f y x or flip f x = \y -> f y x flip f x y = f y x

Aaron Gray

27 Dec 27 Dec

6:22 p.m.

On 24 December 2010 23:58, Daniel Fischer wrote:

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On Saturday 25 December 2010 00:32:38, Aaron Gray wrote:

...
Okay great, works this end too, but what does the 'flip' do ???

It flips the order of arguments to calc. You could also write

main = getContents >>= print . (`calc` []) . lexer

Generally,

flip f = \x y -> f y x

or

flip f x = \y -> f y x

flip f x y = f y x

Thanks Daniel, Aaron

Aaron Gray

6:37 p.m.

On 24 December 2010 23:58, Daniel Fischer wrote:

...

On Saturday 25 December 2010 00:32:38, Aaron Gray wrote:

...
Okay great, works this end too, but what does the 'flip' do ???

It flips the order of arguments to calc. You could also write

main = getContents >>= print . (`calc` []) . lexer

Generally,

flip f = \x y -> f y x

or

flip f x = \y -> f y x

flip f x y = f y x

Or :- calcExpr :: [Token] -> Int calcExpr tokens = calc tokens [] process = print . calcExpr . lexer makes things clearer. Aaron

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Aaron Gray
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instance for (Show ([(String, Int)] -> Int))

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