# Are these solution the Haskell way ?

9 messages
Open this post in threaded view
|

## Are these solution the Haskell way ?

 Hello, For practising pattern matching and recursion I did recreate some commands of Data,list. My re-implementation of ++ : plusplus :: [a] -> [a] -> [a] plusplus [] [] = [] ; plusplus [] (xs) = xs plusplus (xs) [] = xs plusplus (xs) yx = plusplus' (reverse xs) yx plusplus' :: [a] -> [a] -> [a] plusplus' [] (xs) = xs plusplus' (x:xs) yx = plusplus' xs (x:yx) main = print \$ plusplus ["a","b"] ["c","d"] my re-implementation of init : import Data.Maybe -- | The main entry point. init' :: [a] -> Maybe [a] init' [] = Nothing init' [x] = Just [] init' (x:xs) = Just (x:fromMaybe xs (init' xs)) main = print . init' \$ [1,3] my re-implementation of last : -- | The main entry point. last' :: [a] -> Maybe a last' [] = Nothing last' [x] = Just x last' (_:xs) = last' xs main = print . last' \$ [] Now I wonder if these solutions are the haskell way ? if not so, how can I improve them , Roelof --- Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. http://www.avast.com_______________________________________________ Beginners mailing list [hidden email] http://mail.haskell.org/cgi-bin/mailman/listinfo/beginners
Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

Thanks,

If I re-implement it like this :

plusplus :: [a] -> [a] -> [a]
plusplus [] (xs) = xs
plusplus (x:xs) yx = x : xs plusplus yx

main = print \$ ["a","b"] plusplus ["c","d"]

I see this error appear :

src/Main.hs@3:26-3:40
Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with actual type
[a]
Relevant bindings include yx :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) xs :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) x :: a (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) plusplus :: [a] -> [a] -> [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) The function
xs
is applied to two arguments, but its type
[a]
has none

So for me not a aha moment.  I was hoping I would get it

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:27:
Hi Roelof,

If you don't consider it cheating (and I suggest you shouldn't, having had a stab at the answers), you will find great enlightenment looking at how these functions are *actually* implemented in the wild.  Did you know that there is a "source" link for each function on Hackage?  By clicking on that, for your first example, you can compare the actual implementation of (++) with your "plusplus" function:

(By the way, it's that function in particular that gave me one of my first "Aha!" moments in Haskell... it's quite beautiful in its simplicity).

One thing with viewing the source on Hackage is that sometimes it can be a little more confusing than it needs to be for the sake of efficiency.  A really good source for good, readable examples of Prelude functions in Haskell is the Haskell Report:

(In this case, though, the implementation is the same).

Having a shot at defining these library functions yourself, as you have done, and then comparing your version with the "official" version in the prelude is a great way to learn good style!

-Dani.

2015-05-13 15:10 GMT+09:00 Roelof Wobben :
Hello,

For practising pattern matching and recursion I did recreate some commands of Data,list.

My re-implementation of ++ :

plusplus :: [a] -> [a] -> [a]
plusplus [] [] = [] ;
plusplus [] (xs) = xs
plusplus (xs) [] = xs
plusplus (xs) yx = plusplus' (reverse xs) yx

plusplus' :: [a] -> [a] -> [a]
plusplus' [] (xs) = xs
plusplus' (x:xs) yx = plusplus' xs (x:yx)

main = print \$ plusplus ["a","b"] ["c","d"]

my re-implementation of init :

import Data.Maybe

-- | The main entry point.
init' :: [a] -> Maybe [a]
init' [] = Nothing
init' [x] = Just []
init' (x:xs) = Just (x:fromMaybe xs (init' xs))

main = print . init' \$ [1,3]

my re-implementation of last :

-- | The main entry point.
last' :: [a] -> Maybe a
last' [] = Nothing
last' [x] = Just x
last' (_:xs) = last' xs

main = print . last' \$ []

Now I wonder if these solutions are the haskell way ? if not so, how can I improve them ,

Roelof

---
Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware.
http://www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]
Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

Ah, that's just a small syntactic issue -- in Haskell, operators are infix (go between the arguments) by default, but named functions are not.  So you would have to write it:

plusplus [] xs = xs
plusplus (x:xs) ys = x : plusplus xs ys

2015-05-13 15:39 GMT+09:00 Roelof Wobben :
Thanks,

If I re-implement it like this :

plusplus :: [a] -> [a] -> [a]
plusplus [] (xs) = xs
plusplus (x:xs) yx = x : xs plusplus yx

main = print \$ ["a","b"] plusplus ["c","d"]

I see this error appear :

src/Main.hs@3:26-3:40
Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with actual type
[a]
Relevant bindings include yx :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) xs :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) x :: a (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) plusplus :: [a] -> [a] -> [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) The function
xs
is applied to two arguments, but its type
[a]
has none

So for me not a aha moment.  I was hoping I would get it

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:27:
Hi Roelof,

If you don't consider it cheating (and I suggest you shouldn't, having had a stab at the answers), you will find great enlightenment looking at how these functions are *actually* implemented in the wild.  Did you know that there is a "source" link for each function on Hackage?  By clicking on that, for your first example, you can compare the actual implementation of (++) with your "plusplus" function:

(By the way, it's that function in particular that gave me one of my first "Aha!" moments in Haskell... it's quite beautiful in its simplicity).

One thing with viewing the source on Hackage is that sometimes it can be a little more confusing than it needs to be for the sake of efficiency.  A really good source for good, readable examples of Prelude functions in Haskell is the Haskell Report:

(In this case, though, the implementation is the same).

Having a shot at defining these library functions yourself, as you have done, and then comparing your version with the "official" version in the prelude is a great way to learn good style!

-Dani.

2015-05-13 15:10 GMT+09:00 Roelof Wobben :
Hello,

For practising pattern matching and recursion I did recreate some commands of Data,list.

My re-implementation of ++ :

plusplus :: [a] -> [a] -> [a]
plusplus [] [] = [] ;
plusplus [] (xs) = xs
plusplus (xs) [] = xs
plusplus (xs) yx = plusplus' (reverse xs) yx

plusplus' :: [a] -> [a] -> [a]
plusplus' [] (xs) = xs
plusplus' (x:xs) yx = plusplus' xs (x:yx)

main = print \$ plusplus ["a","b"] ["c","d"]

my re-implementation of init :

import Data.Maybe

-- | The main entry point.
init' :: [a] -> Maybe [a]
init' [] = Nothing
init' [x] = Just []
init' (x:xs) = Just (x:fromMaybe xs (init' xs))

main = print . init' \$ [1,3]

my re-implementation of last :

-- | The main entry point.
last' :: [a] -> Maybe a
last' [] = Nothing
last' [x] = Just x
last' (_:xs) = last' xs

main = print . last' \$ []

Now I wonder if these solutions are the haskell way ? if not so, how can I improve them ,

Roelof

---
Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware.
http://www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

_______________________________________________
Beginners mailing list
[hidden email]
Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

Thanks,

So the answer is x and the rest of xs and ys.
How do x then get added to ys.

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:52:
Ah, that's just a small syntactic issue -- in Haskell, operators are infix (go between the arguments) by default, but named functions are not.  So you would have to write it:

plusplus [] xs = xs
plusplus (x:xs) ys = x : plusplus xs ys

2015-05-13 15:39 GMT+09:00 Roelof Wobben :
Thanks,

If I re-implement it like this :

plusplus :: [a] -> [a] -> [a]
plusplus [] (xs) = xs
plusplus (x:xs) yx = x : xs plusplus yx

main = print \$ ["a","b"] plusplus ["c","d"]

I see this error appear :

src/Main.hs@3:26-3:40
Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with actual type
[a]
Relevant bindings include yx :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) xs :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) x :: a (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) plusplus :: [a] -> [a] -> [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) The function
xs
is applied to two arguments, but its type
[a]
has none

So for me not a aha moment.  I was hoping I would get it

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:27:
Hi Roelof,

If you don't consider it cheating (and I suggest you shouldn't, having had a stab at the answers), you will find great enlightenment looking at how these functions are *actually* implemented in the wild.  Did you know that there is a "source" link for each function on Hackage?  By clicking on that, for your first example, you can compare the actual implementation of (++) with your "plusplus" function:

(By the way, it's that function in particular that gave me one of my first "Aha!" moments in Haskell... it's quite beautiful in its simplicity).

One thing with viewing the source on Hackage is that sometimes it can be a little more confusing than it needs to be for the sake of efficiency.  A really good source for good, readable examples of Prelude functions in Haskell is the Haskell Report:

(In this case, though, the implementation is the same).

Having a shot at defining these library functions yourself, as you have done, and then comparing your version with the "official" version in the prelude is a great way to learn good style!

-Dani.

2015-05-13 15:10 GMT+09:00 Roelof Wobben :
Hello,

For practising pattern matching and recursion I did recreate some commands of Data,list.

My re-implementation of ++ :

plusplus :: [a] -> [a] -> [a]
plusplus [] [] = [] ;
plusplus [] (xs) = xs
plusplus (xs) [] = xs
plusplus (xs) yx = plusplus' (reverse xs) yx

plusplus' :: [a] -> [a] -> [a]
plusplus' [] (xs) = xs
plusplus' (x:xs) yx = plusplus' xs (x:yx)

main = print \$ plusplus ["a","b"] ["c","d"]

my re-implementation of init :

import Data.Maybe

-- | The main entry point.
init' :: [a] -> Maybe [a]
init' [] = Nothing
init' [x] = Just []
init' (x:xs) = Just (x:fromMaybe xs (init' xs))

main = print . init' \$ [1,3]

my re-implementation of last :

-- | The main entry point.
last' :: [a] -> Maybe a
last' [] = Nothing
last' [x] = Just x
last' (_:xs) = last' xs

main = print . last' \$ []

Now I wonder if these solutions are the haskell way ? if not so, how can I improve them ,

Roelof

---
Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware.
http://www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]
Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

Think about what the possible values of "xs" might be, and trace through the next call to "plusplus xs ys".

It would help if I didn't name the variables stupidly... here is a slightly better version:

plusplus [] ys = ys
plusplus (x:xs) ys = x : plusplus xs ys

If it helps, try tracing through the steps required to evaluate "plusplus [1, 2] [3, 4]" manually (using a pen and paper, not on the computer)

2015-05-13 15:55 GMT+09:00 Roelof Wobben :
Thanks,

So the answer is x and the rest of xs and ys.
How do x then get added to ys.

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:52:
Ah, that's just a small syntactic issue -- in Haskell, operators are infix (go between the arguments) by default, but named functions are not.  So you would have to write it:

plusplus [] xs = xs
plusplus (x:xs) ys = x : plusplus xs ys

2015-05-13 15:39 GMT+09:00 Roelof Wobben :
Thanks,

If I re-implement it like this :

plusplus :: [a] -> [a] -> [a]
plusplus [] (xs) = xs
plusplus (x:xs) yx = x : xs plusplus yx

main = print \$ ["a","b"] plusplus ["c","d"]

I see this error appear :

src/Main.hs@3:26-3:40
Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with actual type
[a]
Relevant bindings include yx :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) xs :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) x :: a (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) plusplus :: [a] -> [a] -> [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) The function
xs
is applied to two arguments, but its type
[a]
has none

So for me not a aha moment.  I was hoping I would get it

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:27:
Hi Roelof,

If you don't consider it cheating (and I suggest you shouldn't, having had a stab at the answers), you will find great enlightenment looking at how these functions are *actually* implemented in the wild.  Did you know that there is a "source" link for each function on Hackage?  By clicking on that, for your first example, you can compare the actual implementation of (++) with your "plusplus" function:

(By the way, it's that function in particular that gave me one of my first "Aha!" moments in Haskell... it's quite beautiful in its simplicity).

One thing with viewing the source on Hackage is that sometimes it can be a little more confusing than it needs to be for the sake of efficiency.  A really good source for good, readable examples of Prelude functions in Haskell is the Haskell Report:

(In this case, though, the implementation is the same).

Having a shot at defining these library functions yourself, as you have done, and then comparing your version with the "official" version in the prelude is a great way to learn good style!

-Dani.

2015-05-13 15:10 GMT+09:00 Roelof Wobben :
Hello,

For practising pattern matching and recursion I did recreate some commands of Data,list.

My re-implementation of ++ :

plusplus :: [a] -> [a] -> [a]
plusplus [] [] = [] ;
plusplus [] (xs) = xs
plusplus (xs) [] = xs
plusplus (xs) yx = plusplus' (reverse xs) yx

plusplus' :: [a] -> [a] -> [a]
plusplus' [] (xs) = xs
plusplus' (x:xs) yx = plusplus' xs (x:yx)

main = print \$ plusplus ["a","b"] ["c","d"]

my re-implementation of init :

import Data.Maybe

-- | The main entry point.
init' :: [a] -> Maybe [a]
init' [] = Nothing
init' [x] = Just []
init' (x:xs) = Just (x:fromMaybe xs (init' xs))

main = print . init' \$ [1,3]

my re-implementation of last :

-- | The main entry point.
last' :: [a] -> Maybe a
last' [] = Nothing
last' [x] = Just x
last' (_:xs) = last' xs

main = print . last' \$ []

Now I wonder if these solutions are the haskell way ? if not so, how can I improve them ,

Roelof

---
Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware.
http://www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

_______________________________________________
Beginners mailing list
[hidden email]
Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

In reply to this post by Daniel P. Wright
Hello,

I see it.

plusplus [1,2] [3,4]
plusplus  [1] : plusplus [2] [3,4]
plusplus [1] :  [2] : plusplus [] [3,4]
plusplus [1] : [2] : [3,4]
[1:2:3:4]

I think I need some more practice on this.
Does anyone know a good exercise ?

Roelof

Daniel P. Wright schreef op 13-5-2015 om 9:08:
Think about what the possible values of "xs" might be, and trace through the next call to "plusplus xs ys".

It would help if I didn't name the variables stupidly... here is a slightly better version:

plusplus [] ys = ys
plusplus (x:xs) ys = x : plusplus xs ys

If it helps, try tracing through the steps required to evaluate "plusplus [1, 2] [3, 4]" manually (using a pen and paper, not on the computer)

2015-05-13 15:55 GMT+09:00 Roelof Wobben :
Thanks,

So the answer is x and the rest of xs and ys.
How do x then get added to ys.

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:52:
Ah, that's just a small syntactic issue -- in Haskell, operators are infix (go between the arguments) by default, but named functions are not.  So you would have to write it:

plusplus [] xs = xs
plusplus (x:xs) ys = x : plusplus xs ys

2015-05-13 15:39 GMT+09:00 Roelof Wobben :
Thanks,

If I re-implement it like this :

plusplus :: [a] -> [a] -> [a]
plusplus [] (xs) = xs
plusplus (x:xs) yx = x : xs plusplus yx

main = print \$ ["a","b"] plusplus ["c","d"]

I see this error appear :

src/Main.hs@3:26-3:40
Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with actual type
[a]
Relevant bindings include yx :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) xs :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) x :: a (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) plusplus :: [a] -> [a] -> [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) The function
xs
is applied to two arguments, but its type
[a]
has none

So for me not a aha moment.  I was hoping I would get it

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:27:
Hi Roelof,

If you don't consider it cheating (and I suggest you shouldn't, having had a stab at the answers), you will find great enlightenment looking at how these functions are *actually* implemented in the wild.  Did you know that there is a "source" link for each function on Hackage?  By clicking on that, for your first example, you can compare the actual implementation of (++) with your "plusplus" function:

(By the way, it's that function in particular that gave me one of my first "Aha!" moments in Haskell... it's quite beautiful in its simplicity).

One thing with viewing the source on Hackage is that sometimes it can be a little more confusing than it needs to be for the sake of efficiency.  A really good source for good, readable examples of Prelude functions in Haskell is the Haskell Report:

(In this case, though, the implementation is the same).

Having a shot at defining these library functions yourself, as you have done, and then comparing your version with the "official" version in the prelude is a great way to learn good style!

-Dani.

2015-05-13 15:10 GMT+09:00 Roelof Wobben :
Hello,

For practising pattern matching and recursion I did recreate some commands of Data,list.

My re-implementation of ++ :

plusplus :: [a] -> [a] -> [a]
plusplus [] [] = [] ;
plusplus [] (xs) = xs
plusplus (xs) [] = xs
plusplus (xs) yx = plusplus' (reverse xs) yx

plusplus' :: [a] -> [a] -> [a]
plusplus' [] (xs) = xs
plusplus' (x:xs) yx = plusplus' xs (x:yx)

main = print \$ plusplus ["a","b"] ["c","d"]

my re-implementation of init :

import Data.Maybe

-- | The main entry point.
init' :: [a] -> Maybe [a]
init' [] = Nothing
init' [x] = Just []
init' (x:xs) = Just (x:fromMaybe xs (init' xs))

main = print . init' \$ [1,3]

my re-implementation of last :

-- | The main entry point.
last' :: [a] -> Maybe a
last' [] = Nothing
last' [x] = Just x
last' (_:xs) = last' xs

main = print . last' \$ []

Now I wonder if these solutions are the haskell way ? if not so, how can I improve them ,

Roelof

---
Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware.
http://www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]
Open this post in threaded view
|

## Re: Are these solution the Haskell way ?

The evaluation would go the way Frerich shows it. In particular you should know that these two notations are equivalent.

[1,2,3] == 1 : 2 : 3 : []

The cons (:) operator (with type a -> [a] -> [a]) adds an element to the start of a list. Also, as it is right associative

1 : 2 : 3 : []
==  1 : (2 : (3 : []))   { right associativity }
==  1 : (2 : [3])        { add element to empty list }
==  1 : [2, 3]           { add element }
==  [1, 2, 3]            { add element }

On 13 May 2015 at 12:50, Roelof Wobben wrote:
Hello,

I see it.

plusplus [1,2] [3,4]
plusplus  [1] : plusplus [2] [3,4]
plusplus [1] :  [2] : plusplus [] [3,4]
plusplus [1] : [2] : [3,4]
[1:2:3:4]

I think I need some more practice on this.
Does anyone know a good exercise ?

Roelof

Daniel P. Wright schreef op 13-5-2015 om 9:08:
Think about what the possible values of "xs" might be, and trace through the next call to "plusplus xs ys".

It would help if I didn't name the variables stupidly... here is a slightly better version:

plusplus [] ys = ys
plusplus (x:xs) ys = x : plusplus xs ys

If it helps, try tracing through the steps required to evaluate "plusplus [1, 2] [3, 4]" manually (using a pen and paper, not on the computer)

2015-05-13 15:55 GMT+09:00 Roelof Wobben :
Thanks,

So the answer is x and the rest of xs and ys.
How do x then get added to ys.

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:52:
Ah, that's just a small syntactic issue -- in Haskell, operators are infix (go between the arguments) by default, but named functions are not.  So you would have to write it:

plusplus [] xs = xs
plusplus (x:xs) ys = x : plusplus xs ys

2015-05-13 15:39 GMT+09:00 Roelof Wobben :
Thanks,

If I re-implement it like this :

plusplus :: [a] -> [a] -> [a]
plusplus [] (xs) = xs
plusplus (x:xs) yx = x : xs plusplus yx

main = print \$ ["a","b"] plusplus ["c","d"]

I see this error appear :

src/Main.hs@3:26-3:40
Couldn't match expected type ‘([a0] -> [a0] -> [a0]) -> [a] -> [a]’ with actual type
[a]
Relevant bindings include yx :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:17) xs :: [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:13) x :: a (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:3:11) plusplus :: [a] -> [a] -> [a] (bound at /home/app/isolation-runner-work/projects/112825/session.207/src/src/Main.hs:2:1) The function
xs
is applied to two arguments, but its type
[a]
has none

So for me not a aha moment.  I was hoping I would get it

Roelof

Daniel P. Wright schreef op 13-5-2015 om 8:27:
Hi Roelof,

If you don't consider it cheating (and I suggest you shouldn't, having had a stab at the answers), you will find great enlightenment looking at how these functions are *actually* implemented in the wild.  Did you know that there is a "source" link for each function on Hackage?  By clicking on that, for your first example, you can compare the actual implementation of (++) with your "plusplus" function:

(By the way, it's that function in particular that gave me one of my first "Aha!" moments in Haskell... it's quite beautiful in its simplicity).

One thing with viewing the source on Hackage is that sometimes it can be a little more confusing than it needs to be for the sake of efficiency.  A really good source for good, readable examples of Prelude functions in Haskell is the Haskell Report:

(In this case, though, the implementation is the same).

Having a shot at defining these library functions yourself, as you have done, and then comparing your version with the "official" version in the prelude is a great way to learn good style!

-Dani.

2015-05-13 15:10 GMT+09:00 Roelof Wobben :
Hello,

For practising pattern matching and recursion I did recreate some commands of Data,list.

My re-implementation of ++ :

plusplus :: [a] -> [a] -> [a]
plusplus [] [] = [] ;
plusplus [] (xs) = xs
plusplus (xs) [] = xs
plusplus (xs) yx = plusplus' (reverse xs) yx

plusplus' :: [a] -> [a] -> [a]
plusplus' [] (xs) = xs
plusplus' (x:xs) yx = plusplus' xs (x:yx)

main = print \$ plusplus ["a","b"] ["c","d"]

my re-implementation of init :

import Data.Maybe

-- | The main entry point.
init' :: [a] -> Maybe [a]
init' [] = Nothing
init' [x] = Just []
init' (x:xs) = Just (x:fromMaybe xs (init' xs))

main = print . init' \$ [1,3]

my re-implementation of last :

-- | The main entry point.
last' :: [a] -> Maybe a
last' [] = Nothing
last' [x] = Just x
last' (_:xs) = last' xs

main = print . last' \$ []

Now I wonder if these solutions are the haskell way ? if not so, how can I improve them ,

Roelof

---
Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware.
http://www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]

```_______________________________________________
Beginners mailing list
[hidden email]
```

 Dit e-mailbericht is gecontroleerd op virussen met Avast antivirussoftware. www.avast.com

_______________________________________________
Beginners mailing list
[hidden email]