F# function, how it really works

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I got function like this one.



let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
let x = c‐6


That's what I understand so far:
"fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
"let d" is kind a function that takes "a" and "b" and returns the result.
But if somebody could explain what exactly happened in:



let d = a 10 in d ‐ 4*b









share|improve this question



























    2















    I got function like this one.



    let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
    let x = c‐6


    That's what I understand so far:
    "fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
    "let d" is kind a function that takes "a" and "b" and returns the result.
    But if somebody could explain what exactly happened in:



    let d = a 10 in d ‐ 4*b









    share|improve this question

























      2












      2








      2








      I got function like this one.



      let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
      let x = c‐6


      That's what I understand so far:
      "fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
      "let d" is kind a function that takes "a" and "b" and returns the result.
      But if somebody could explain what exactly happened in:



      let d = a 10 in d ‐ 4*b









      share|improve this question














      I got function like this one.



      let c = (fun a b ‐> let d = a 10 in d ‐ 4*b) (fun c ‐> c + c) 5
      let x = c‐6


      That's what I understand so far:
      "fun c ‐> c + c" is first argument for "fun a b" and "5" is second.
      "let d" is kind a function that takes "a" and "b" and returns the result.
      But if somebody could explain what exactly happened in:



      let d = a 10 in d ‐ 4*b






      f#






      share|improve this question













      share|improve this question











      share|improve this question




      share|improve this question










      asked Dec 31 '18 at 12:07









      silvanestisilvanesti

      133




      133
























          1 Answer
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          active

          oldest

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          1














          in is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:



          fun a b =
          let d = a 10
          d - 4 * b


          a is invoked with 10 as argument and the results is assigned to d. Later on d - 4 * b is calculated and returned from the function.



          For the example you have a is fun c -> c + c so invoking it with 10 returns 20.



          20 - 4 * 5 = 0 so c is set to 0 and x will be 0 - 6 = -6.



          From F# interactive:



          > let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
          val c : int = 0





          share|improve this answer
























          • I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

            – silvanesti
            Dec 31 '18 at 12:57











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          1 Answer
          1






          active

          oldest

          votes








          1 Answer
          1






          active

          oldest

          votes









          active

          oldest

          votes






          active

          oldest

          votes









          1














          in is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:



          fun a b =
          let d = a 10
          d - 4 * b


          a is invoked with 10 as argument and the results is assigned to d. Later on d - 4 * b is calculated and returned from the function.



          For the example you have a is fun c -> c + c so invoking it with 10 returns 20.



          20 - 4 * 5 = 0 so c is set to 0 and x will be 0 - 6 = -6.



          From F# interactive:



          > let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
          val c : int = 0





          share|improve this answer
























          • I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

            – silvanesti
            Dec 31 '18 at 12:57
















          1














          in is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:



          fun a b =
          let d = a 10
          d - 4 * b


          a is invoked with 10 as argument and the results is assigned to d. Later on d - 4 * b is calculated and returned from the function.



          For the example you have a is fun c -> c + c so invoking it with 10 returns 20.



          20 - 4 * 5 = 0 so c is set to 0 and x will be 0 - 6 = -6.



          From F# interactive:



          > let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
          val c : int = 0





          share|improve this answer
























          • I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

            – silvanesti
            Dec 31 '18 at 12:57














          1












          1








          1







          in is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:



          fun a b =
          let d = a 10
          d - 4 * b


          a is invoked with 10 as argument and the results is assigned to d. Later on d - 4 * b is calculated and returned from the function.



          For the example you have a is fun c -> c + c so invoking it with 10 returns 20.



          20 - 4 * 5 = 0 so c is set to 0 and x will be 0 - 6 = -6.



          From F# interactive:



          > let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
          val c : int = 0





          share|improve this answer













          in is used here as part of verbose syntax. You can rewrite it in leightweight syntax to make it a bit more readable. It would be something like that:



          fun a b =
          let d = a 10
          d - 4 * b


          a is invoked with 10 as argument and the results is assigned to d. Later on d - 4 * b is calculated and returned from the function.



          For the example you have a is fun c -> c + c so invoking it with 10 returns 20.



          20 - 4 * 5 = 0 so c is set to 0 and x will be 0 - 6 = -6.



          From F# interactive:



          > let c = (fun a b -> let d = a 10 in d - 4*b) (fun c -> c + c) 5;;
          val c : int = 0






          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Dec 31 '18 at 12:16









          MarcinJuraszekMarcinJuraszek

          108k10138215




          108k10138215













          • I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

            – silvanesti
            Dec 31 '18 at 12:57



















          • I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

            – silvanesti
            Dec 31 '18 at 12:57

















          I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

          – silvanesti
          Dec 31 '18 at 12:57





          I did not consider the possibility that "a" could be a function and takes "10" as an argument. Thanks, that really helped me.

          – silvanesti
          Dec 31 '18 at 12:57




















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