Re: [Haskell-beginners] Beginners Digest, Vol 161, Issue 1

The important thing here, is that print is a function from a "pure" value (any type implements the Show typeclass. (similar to trait in rust, or interface in some languages)) and creates a "IO" value, which can be thought of as a IO action. inside the IO monad, that just means running it. ‪On Sun, Dec 12, 2021 at 7:05 PM ‫יהושע ולך‬‎ wrote:‬

Notice: (from ghci)

Prelude GHC.Stack> :t callStack callStack :: CallStack Prelude GHC.Stack> :t getCallStack getCallStack :: CallStack -> [([Char], SrcLoc)] Prelude GHC.Stack> :t getCallStack callStack getCallStack callStack :: [([Char], SrcLoc)] Prelude GHC.Stack> :t length (getCallStack callStack) length (getCallStack callStack) :: Int

Prelude GHC.Stack> :t print (length (getCallStack callStack)) print (length (getCallStack callStack)) :: IO ()

Prelude GHC.Stack> :t print print :: Show a => a -> IO ()

the `print` function takes a pure value, and creates an IO value from it. (actually printing is an IO action. the value it prints is not)

On Sun, Dec 12, 2021 at 5:32 PM Michael Turner < michael.eugene.turner@gmail.com> wrote:

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Regards, Michael Turner Executive Director Project Persephone 1-25-33 Takadanobaba

Shinjuku-ku Tokyo 169-0075 Mobile: +81 (90) 5203-8682 turner@projectpersephone.org

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On Sun, Dec 12, 2021 at 8:48 PM wrote:

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have another look at

https://hackage.haskell.org/packaggetCallStackgetCallStacke/base-4.16.0.0/do...

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< https://hackage.haskell.org/package/base-4.16.0.0/docs/GHC-Stack.html#v:call...

I'd been looking at that already, over and over. Still not seeing it. But your proposed fix does work. Thank you.

...

let l = getCallStack callStack print (length l) ``` the `getCallStack` does not return an "IO" value..

And yet it's it's somehow an IO value in "print (length (getCallStack callStack))"?

I just don't get this.

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On Sun, Dec 12, 2021, 12:58 Michael Turner < michael.eugene.turner@gmail.com> wrote:

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The following prints 1, as you'd expect: ----------------- import GHC.Stack

foo :: HasCallStack => IO () foo = do print (length (getCallStack callStack))

main = foo ------------------- But when I make it this: ... l <- getCallStack callStack ; print (length l)

I get all this: ------------------------------------------ ... • Couldn't match type ‘[]’ with ‘IO’ Expected type: IO ([Char], SrcLoc) Actual type: [([Char], SrcLoc)] • In a stmt of a 'do' block: l <- getCallStack callStack In the expression: do l <- getCallStack callStack print (length l) In an equation for ‘foo’: foo = do l <- getCallStack callStack print (length l) | 5 | l <- getCallStack callStack ; | ^^^^^^^^^^^^^^^^^^^^^^ --------------------------------------------

What am I not seeing?

Regards, Michael Turner Executive Director Project Persephone 1-25-33 Takadanobaba

Shinjuku-ku Tokyo 169-0075 Mobile: +81 (90) 5203-8682 turner@projectpersephone.org

Understand - http://www.projectpersephone.org/ Join - http://www.facebook.com/groups/ProjectPersephone/ Donate - http://www.patreon.com/ProjectPersephone Volunteer - https://github.com/ProjectPersephone

"Love does not consist in gazing at each other, but in looking outward together in the same direction." -- Antoine de Saint-Exupéry

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