Please how do I make this function Tail-recursive in F#?












-1














let ListtoTuple (lst:'a list) :('a * 'a) list =



let rec loop (lt :'a list) acc =

    match lt with         

    | x:: y :: t-> (x,y):: loop  t acc 

    |   _   -> acc

loop lst


Question: ListtoTuple 6 [1..1000000]



I want this kind of result : [(1, 2); (3, 4); (5, 6); (7, 8); (9, 10)...] but I keep getting process is terminated due to StackOverflow. Please, I would like to know if there is anything I am doing wrong.






f# f#-interactive f#-data f#-3.0







share|improve this question
























  • Don't use unrelated tags. It doesn't improve visibility, it only annoys people interested in those tags. As for tail-recursive it means the recursive call comes last.
    – Panagiotis Kanavos
    Nov 12 at 16:43
















-1














let ListtoTuple (lst:'a list) :('a * 'a) list =



let rec loop (lt :'a list) acc =

    match lt with         

    | x:: y :: t-> (x,y):: loop  t acc 

    |   _   -> acc

loop lst


Question: ListtoTuple 6 [1..1000000]



I want this kind of result : [(1, 2); (3, 4); (5, 6); (7, 8); (9, 10)...] but I keep getting process is terminated due to StackOverflow. Please, I would like to know if there is anything I am doing wrong.






f# f#-interactive f#-data f#-3.0







share|improve this question
























  • Don't use unrelated tags. It doesn't improve visibility, it only annoys people interested in those tags. As for tail-recursive it means the recursive call comes last.
    – Panagiotis Kanavos
    Nov 12 at 16:43














-1












-1








-1







let ListtoTuple (lst:'a list) :('a * 'a) list =



let rec loop (lt :'a list) acc =

    match lt with         

    | x:: y :: t-> (x,y):: loop  t acc 

    |   _   -> acc

loop lst


Question: ListtoTuple 6 [1..1000000]



I want this kind of result : [(1, 2); (3, 4); (5, 6); (7, 8); (9, 10)...] but I keep getting process is terminated due to StackOverflow. Please, I would like to know if there is anything I am doing wrong.






f# f#-interactive f#-data f#-3.0







share|improve this question















let ListtoTuple (lst:'a list) :('a * 'a) list =



let rec loop (lt :'a list) acc =

    match lt with         

    | x:: y :: t-> (x,y):: loop  t acc 

    |   _   -> acc

loop lst


Question: ListtoTuple 6 [1..1000000]



I want this kind of result : [(1, 2); (3, 4); (5, 6); (7, 8); (9, 10)...] but I keep getting process is terminated due to StackOverflow. Please, I would like to know if there is anything I am doing wrong.






f# f#-interactive f#-data f#-3.0




f# f#-interactive f#-data f#-3.0






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 12 at 16:56

























asked Nov 12 at 16:40









T.M

389




389












  • Don't use unrelated tags. It doesn't improve visibility, it only annoys people interested in those tags. As for tail-recursive it means the recursive call comes last.
    – Panagiotis Kanavos
    Nov 12 at 16:43


















  • Don't use unrelated tags. It doesn't improve visibility, it only annoys people interested in those tags. As for tail-recursive it means the recursive call comes last.
    – Panagiotis Kanavos
    Nov 12 at 16:43
















Don't use unrelated tags. It doesn't improve visibility, it only annoys people interested in those tags. As for tail-recursive it means the recursive call comes last.
– Panagiotis Kanavos
Nov 12 at 16:43




Don't use unrelated tags. It doesn't improve visibility, it only annoys people interested in those tags. As for tail-recursive it means the recursive call comes last.
– Panagiotis Kanavos
Nov 12 at 16:43












2 Answers
2






active

oldest

votes


















2














Your code is almost correct. The problem is here:



(x,y):: loop f t acc


You are concatenating to the result of loop which means it is not tail recursive because it has to wait for the result of loop to then concatenate.



The key is in the parameter acc which stands for accumulator. That means that is where you need to be concatenating your resulting list, which then gets passed to the next level until there is nothing more to add and then the acc has the completed list which is returned here:



|   _   -> acc





share|improve this answer





















  • Thanks for the explanation AMieres
    – T.M
    Nov 12 at 17:17





















1














From this thread:



let listToPairList lst =

    let rec aux acc lst = 

        match lst with

        |          -> acc |> List.rev

        | x::      -> (x,x)::acc |> List.rev

        | x1::x2::xs -> aux ((x1,x2)::acc) xs

    aux  lst





share|improve this answer





















  • Thank you gileCAD
    – T.M
    Nov 12 at 17:16











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









2














Your code is almost correct. The problem is here:



(x,y):: loop f t acc


You are concatenating to the result of loop which means it is not tail recursive because it has to wait for the result of loop to then concatenate.



The key is in the parameter acc which stands for accumulator. That means that is where you need to be concatenating your resulting list, which then gets passed to the next level until there is nothing more to add and then the acc has the completed list which is returned here:



|   _   -> acc





share|improve this answer





















  • Thanks for the explanation AMieres
    – T.M
    Nov 12 at 17:17


















2














Your code is almost correct. The problem is here:



(x,y):: loop f t acc


You are concatenating to the result of loop which means it is not tail recursive because it has to wait for the result of loop to then concatenate.



The key is in the parameter acc which stands for accumulator. That means that is where you need to be concatenating your resulting list, which then gets passed to the next level until there is nothing more to add and then the acc has the completed list which is returned here:



|   _   -> acc





share|improve this answer





















  • Thanks for the explanation AMieres
    – T.M
    Nov 12 at 17:17
















2












2








2






Your code is almost correct. The problem is here:



(x,y):: loop f t acc


You are concatenating to the result of loop which means it is not tail recursive because it has to wait for the result of loop to then concatenate.



The key is in the parameter acc which stands for accumulator. That means that is where you need to be concatenating your resulting list, which then gets passed to the next level until there is nothing more to add and then the acc has the completed list which is returned here:



|   _   -> acc





share|improve this answer












Your code is almost correct. The problem is here:



(x,y):: loop f t acc


You are concatenating to the result of loop which means it is not tail recursive because it has to wait for the result of loop to then concatenate.



The key is in the parameter acc which stands for accumulator. That means that is where you need to be concatenating your resulting list, which then gets passed to the next level until there is nothing more to add and then the acc has the completed list which is returned here:



|   _   -> acc






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 at 16:57









AMieres

2,050510




2,050510












  • Thanks for the explanation AMieres
    – T.M
    Nov 12 at 17:17




















  • Thanks for the explanation AMieres
    – T.M
    Nov 12 at 17:17


















Thanks for the explanation AMieres
– T.M
Nov 12 at 17:17






Thanks for the explanation AMieres
– T.M
Nov 12 at 17:17















1














From this thread:



let listToPairList lst =

    let rec aux acc lst = 

        match lst with

        |          -> acc |> List.rev

        | x::      -> (x,x)::acc |> List.rev

        | x1::x2::xs -> aux ((x1,x2)::acc) xs

    aux  lst





share|improve this answer





















  • Thank you gileCAD
    – T.M
    Nov 12 at 17:16
















1














From this thread:



let listToPairList lst =

    let rec aux acc lst = 

        match lst with

        |          -> acc |> List.rev

        | x::      -> (x,x)::acc |> List.rev

        | x1::x2::xs -> aux ((x1,x2)::acc) xs

    aux  lst





share|improve this answer





















  • Thank you gileCAD
    – T.M
    Nov 12 at 17:16














1












1








1






From this thread:



let listToPairList lst =

    let rec aux acc lst = 

        match lst with

        |          -> acc |> List.rev

        | x::      -> (x,x)::acc |> List.rev

        | x1::x2::xs -> aux ((x1,x2)::acc) xs

    aux  lst





share|improve this answer












From this thread:



let listToPairList lst =

    let rec aux acc lst = 

        match lst with

        |          -> acc |> List.rev

        | x::      -> (x,x)::acc |> List.rev

        | x1::x2::xs -> aux ((x1,x2)::acc) xs

    aux  lst






share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 at 16:55









gileCAD

1,31256




1,31256












  • Thank you gileCAD
    – T.M
    Nov 12 at 17:16


















  • Thank you gileCAD
    – T.M
    Nov 12 at 17:16
















Thank you gileCAD
– T.M
Nov 12 at 17:16




Thank you gileCAD
– T.M
Nov 12 at 17:16


















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