Does using currying result in lower performance in F#?











up vote
11
down vote

favorite
1












When writing a function that can accept currying, you can write it as a single-argument function that returns a function. For example,



let add x =
let inner y = x + y
inner


So you can either do:



add 3 4


or:



let add3 = add 3
add3 4


My question, is because you return a function, you are conceptually calling a function twice (the outer function and the inner function). Is this slower than:



let add x y = x + y


or does the compiler optimise invocations of add 3 4 in the curried definition?










share|improve this question
























  • Oops! I was typing from memory. I will edit
    – Cthutu
    Nov 28 at 20:37















up vote
11
down vote

favorite
1












When writing a function that can accept currying, you can write it as a single-argument function that returns a function. For example,



let add x =
let inner y = x + y
inner


So you can either do:



add 3 4


or:



let add3 = add 3
add3 4


My question, is because you return a function, you are conceptually calling a function twice (the outer function and the inner function). Is this slower than:



let add x y = x + y


or does the compiler optimise invocations of add 3 4 in the curried definition?










share|improve this question
























  • Oops! I was typing from memory. I will edit
    – Cthutu
    Nov 28 at 20:37













up vote
11
down vote

favorite
1









up vote
11
down vote

favorite
1






1





When writing a function that can accept currying, you can write it as a single-argument function that returns a function. For example,



let add x =
let inner y = x + y
inner


So you can either do:



add 3 4


or:



let add3 = add 3
add3 4


My question, is because you return a function, you are conceptually calling a function twice (the outer function and the inner function). Is this slower than:



let add x y = x + y


or does the compiler optimise invocations of add 3 4 in the curried definition?










share|improve this question















When writing a function that can accept currying, you can write it as a single-argument function that returns a function. For example,



let add x =
let inner y = x + y
inner


So you can either do:



add 3 4


or:



let add3 = add 3
add3 4


My question, is because you return a function, you are conceptually calling a function twice (the outer function and the inner function). Is this slower than:



let add x y = x + y


or does the compiler optimise invocations of add 3 4 in the curried definition?







f# currying






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Nov 28 at 20:37

























asked Nov 28 at 19:59









Cthutu

5,46652445




5,46652445












  • Oops! I was typing from memory. I will edit
    – Cthutu
    Nov 28 at 20:37


















  • Oops! I was typing from memory. I will edit
    – Cthutu
    Nov 28 at 20:37
















Oops! I was typing from memory. I will edit
– Cthutu
Nov 28 at 20:37




Oops! I was typing from memory. I will edit
– Cthutu
Nov 28 at 20:37












2 Answers
2






active

oldest

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up vote
11
down vote



accepted










let f x   = fun y -> x + y
let g x y = x + y


Looking at these function definitions in dnSpy for an optimized build reveals them to be:



public static int f(int x, int y)
{
return x + y;
}

public static int g(int x, int y)
{
return x + y;
}


This is not that strange because g is actually a short-hand definition for f which is the general case. In F#-like languages function conceptually always take a single value returning a single value. Values might be functions. This is easier to see if one paranthese the function signature for f and g



val f: int -> int -> int
// Actually is
// val f: int -> (int -> int)
// ie f is a function that takes a single int and returns a function that takes a single int and returns an int.


In order to get F# to execute faster on .NET the physical representation of f in an assembly is:



public static int f(int x, int y)


While this is a more natural representation of the F# function.



public static Func<int, int> f(int x)


Would perform poorly though.



Usually F# is clever enough to avoid the overhead of the abstraction by optimization like above and on invocation. However, there are situations where F# can't optimize for you.



Imagine that you are implementing fold



let rec fold f s vs =
match vs with
| v::vs -> fold f (f s v) vs
| -> s


Here F# can't fully optimize f s v. The reason is that f might have a more complex implementation than above that might return a different function depending on s.



If you look in dnSpy you note that F# are invoking function using InvokeFast but this does an internal test to see if it can be invoked fast. In fold we then do this test for each value even though this is the same function.



This is the reason one might sometimes see fold written like this:



let fold f s vs =
let f = OptimizedClosures.FSharpFunc<_, _, _>.Adapt f
let rec loop s vs =
match vs with
| v::vs -> loop (f.Invoke (s, v)) vs
| -> s
loop s vs


Adapt here tests before the loop if f can indeed be optimized and then returns an efficient adapter. In the general case it might still be a bit slower but then this is what the caller intended.



Note; this potential performance degradation doesn't happen for simple function values like 'T -> 'U. This can always be invoked efficiently.



Hope this helps.






share|improve this answer




























    up vote
    9
    down vote













    I tested this in LINQPad 5.



    When compiler optimizations are turned off, the F# compiler will produce different IL for each snippet. In other words, if there are any optimizations going on, it's left up to the JITter, and it may very well be slower to call the first form.



    However, when compiler optimizations are turned on, both forms produce identical IL outputs in every scenario I could think of to test it. In fact, with both forms, calling:



    add 3 4


    yields the IL equivalent of a hard-coded 7, with the entire function call optimized away:



    ldc.i4.7


    In other words, the F# compiler is pretty thorough when it comes to optimizing logically identical code blocks.



    This is not an exhaustive answer, of course, and there could be some case where they are actually treated differently by the compiler.






    share|improve this answer





















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      2 Answers
      2






      active

      oldest

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      2 Answers
      2






      active

      oldest

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      active

      oldest

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      active

      oldest

      votes








      up vote
      11
      down vote



      accepted










      let f x   = fun y -> x + y
      let g x y = x + y


      Looking at these function definitions in dnSpy for an optimized build reveals them to be:



      public static int f(int x, int y)
      {
      return x + y;
      }

      public static int g(int x, int y)
      {
      return x + y;
      }


      This is not that strange because g is actually a short-hand definition for f which is the general case. In F#-like languages function conceptually always take a single value returning a single value. Values might be functions. This is easier to see if one paranthese the function signature for f and g



      val f: int -> int -> int
      // Actually is
      // val f: int -> (int -> int)
      // ie f is a function that takes a single int and returns a function that takes a single int and returns an int.


      In order to get F# to execute faster on .NET the physical representation of f in an assembly is:



      public static int f(int x, int y)


      While this is a more natural representation of the F# function.



      public static Func<int, int> f(int x)


      Would perform poorly though.



      Usually F# is clever enough to avoid the overhead of the abstraction by optimization like above and on invocation. However, there are situations where F# can't optimize for you.



      Imagine that you are implementing fold



      let rec fold f s vs =
      match vs with
      | v::vs -> fold f (f s v) vs
      | -> s


      Here F# can't fully optimize f s v. The reason is that f might have a more complex implementation than above that might return a different function depending on s.



      If you look in dnSpy you note that F# are invoking function using InvokeFast but this does an internal test to see if it can be invoked fast. In fold we then do this test for each value even though this is the same function.



      This is the reason one might sometimes see fold written like this:



      let fold f s vs =
      let f = OptimizedClosures.FSharpFunc<_, _, _>.Adapt f
      let rec loop s vs =
      match vs with
      | v::vs -> loop (f.Invoke (s, v)) vs
      | -> s
      loop s vs


      Adapt here tests before the loop if f can indeed be optimized and then returns an efficient adapter. In the general case it might still be a bit slower but then this is what the caller intended.



      Note; this potential performance degradation doesn't happen for simple function values like 'T -> 'U. This can always be invoked efficiently.



      Hope this helps.






      share|improve this answer

























        up vote
        11
        down vote



        accepted










        let f x   = fun y -> x + y
        let g x y = x + y


        Looking at these function definitions in dnSpy for an optimized build reveals them to be:



        public static int f(int x, int y)
        {
        return x + y;
        }

        public static int g(int x, int y)
        {
        return x + y;
        }


        This is not that strange because g is actually a short-hand definition for f which is the general case. In F#-like languages function conceptually always take a single value returning a single value. Values might be functions. This is easier to see if one paranthese the function signature for f and g



        val f: int -> int -> int
        // Actually is
        // val f: int -> (int -> int)
        // ie f is a function that takes a single int and returns a function that takes a single int and returns an int.


        In order to get F# to execute faster on .NET the physical representation of f in an assembly is:



        public static int f(int x, int y)


        While this is a more natural representation of the F# function.



        public static Func<int, int> f(int x)


        Would perform poorly though.



        Usually F# is clever enough to avoid the overhead of the abstraction by optimization like above and on invocation. However, there are situations where F# can't optimize for you.



        Imagine that you are implementing fold



        let rec fold f s vs =
        match vs with
        | v::vs -> fold f (f s v) vs
        | -> s


        Here F# can't fully optimize f s v. The reason is that f might have a more complex implementation than above that might return a different function depending on s.



        If you look in dnSpy you note that F# are invoking function using InvokeFast but this does an internal test to see if it can be invoked fast. In fold we then do this test for each value even though this is the same function.



        This is the reason one might sometimes see fold written like this:



        let fold f s vs =
        let f = OptimizedClosures.FSharpFunc<_, _, _>.Adapt f
        let rec loop s vs =
        match vs with
        | v::vs -> loop (f.Invoke (s, v)) vs
        | -> s
        loop s vs


        Adapt here tests before the loop if f can indeed be optimized and then returns an efficient adapter. In the general case it might still be a bit slower but then this is what the caller intended.



        Note; this potential performance degradation doesn't happen for simple function values like 'T -> 'U. This can always be invoked efficiently.



        Hope this helps.






        share|improve this answer























          up vote
          11
          down vote



          accepted







          up vote
          11
          down vote



          accepted






          let f x   = fun y -> x + y
          let g x y = x + y


          Looking at these function definitions in dnSpy for an optimized build reveals them to be:



          public static int f(int x, int y)
          {
          return x + y;
          }

          public static int g(int x, int y)
          {
          return x + y;
          }


          This is not that strange because g is actually a short-hand definition for f which is the general case. In F#-like languages function conceptually always take a single value returning a single value. Values might be functions. This is easier to see if one paranthese the function signature for f and g



          val f: int -> int -> int
          // Actually is
          // val f: int -> (int -> int)
          // ie f is a function that takes a single int and returns a function that takes a single int and returns an int.


          In order to get F# to execute faster on .NET the physical representation of f in an assembly is:



          public static int f(int x, int y)


          While this is a more natural representation of the F# function.



          public static Func<int, int> f(int x)


          Would perform poorly though.



          Usually F# is clever enough to avoid the overhead of the abstraction by optimization like above and on invocation. However, there are situations where F# can't optimize for you.



          Imagine that you are implementing fold



          let rec fold f s vs =
          match vs with
          | v::vs -> fold f (f s v) vs
          | -> s


          Here F# can't fully optimize f s v. The reason is that f might have a more complex implementation than above that might return a different function depending on s.



          If you look in dnSpy you note that F# are invoking function using InvokeFast but this does an internal test to see if it can be invoked fast. In fold we then do this test for each value even though this is the same function.



          This is the reason one might sometimes see fold written like this:



          let fold f s vs =
          let f = OptimizedClosures.FSharpFunc<_, _, _>.Adapt f
          let rec loop s vs =
          match vs with
          | v::vs -> loop (f.Invoke (s, v)) vs
          | -> s
          loop s vs


          Adapt here tests before the loop if f can indeed be optimized and then returns an efficient adapter. In the general case it might still be a bit slower but then this is what the caller intended.



          Note; this potential performance degradation doesn't happen for simple function values like 'T -> 'U. This can always be invoked efficiently.



          Hope this helps.






          share|improve this answer












          let f x   = fun y -> x + y
          let g x y = x + y


          Looking at these function definitions in dnSpy for an optimized build reveals them to be:



          public static int f(int x, int y)
          {
          return x + y;
          }

          public static int g(int x, int y)
          {
          return x + y;
          }


          This is not that strange because g is actually a short-hand definition for f which is the general case. In F#-like languages function conceptually always take a single value returning a single value. Values might be functions. This is easier to see if one paranthese the function signature for f and g



          val f: int -> int -> int
          // Actually is
          // val f: int -> (int -> int)
          // ie f is a function that takes a single int and returns a function that takes a single int and returns an int.


          In order to get F# to execute faster on .NET the physical representation of f in an assembly is:



          public static int f(int x, int y)


          While this is a more natural representation of the F# function.



          public static Func<int, int> f(int x)


          Would perform poorly though.



          Usually F# is clever enough to avoid the overhead of the abstraction by optimization like above and on invocation. However, there are situations where F# can't optimize for you.



          Imagine that you are implementing fold



          let rec fold f s vs =
          match vs with
          | v::vs -> fold f (f s v) vs
          | -> s


          Here F# can't fully optimize f s v. The reason is that f might have a more complex implementation than above that might return a different function depending on s.



          If you look in dnSpy you note that F# are invoking function using InvokeFast but this does an internal test to see if it can be invoked fast. In fold we then do this test for each value even though this is the same function.



          This is the reason one might sometimes see fold written like this:



          let fold f s vs =
          let f = OptimizedClosures.FSharpFunc<_, _, _>.Adapt f
          let rec loop s vs =
          match vs with
          | v::vs -> loop (f.Invoke (s, v)) vs
          | -> s
          loop s vs


          Adapt here tests before the loop if f can indeed be optimized and then returns an efficient adapter. In the general case it might still be a bit slower but then this is what the caller intended.



          Note; this potential performance degradation doesn't happen for simple function values like 'T -> 'U. This can always be invoked efficiently.



          Hope this helps.







          share|improve this answer












          share|improve this answer



          share|improve this answer










          answered Nov 28 at 21:18









          Just another metaprogrammer

          8,62522141




          8,62522141
























              up vote
              9
              down vote













              I tested this in LINQPad 5.



              When compiler optimizations are turned off, the F# compiler will produce different IL for each snippet. In other words, if there are any optimizations going on, it's left up to the JITter, and it may very well be slower to call the first form.



              However, when compiler optimizations are turned on, both forms produce identical IL outputs in every scenario I could think of to test it. In fact, with both forms, calling:



              add 3 4


              yields the IL equivalent of a hard-coded 7, with the entire function call optimized away:



              ldc.i4.7


              In other words, the F# compiler is pretty thorough when it comes to optimizing logically identical code blocks.



              This is not an exhaustive answer, of course, and there could be some case where they are actually treated differently by the compiler.






              share|improve this answer

























                up vote
                9
                down vote













                I tested this in LINQPad 5.



                When compiler optimizations are turned off, the F# compiler will produce different IL for each snippet. In other words, if there are any optimizations going on, it's left up to the JITter, and it may very well be slower to call the first form.



                However, when compiler optimizations are turned on, both forms produce identical IL outputs in every scenario I could think of to test it. In fact, with both forms, calling:



                add 3 4


                yields the IL equivalent of a hard-coded 7, with the entire function call optimized away:



                ldc.i4.7


                In other words, the F# compiler is pretty thorough when it comes to optimizing logically identical code blocks.



                This is not an exhaustive answer, of course, and there could be some case where they are actually treated differently by the compiler.






                share|improve this answer























                  up vote
                  9
                  down vote










                  up vote
                  9
                  down vote









                  I tested this in LINQPad 5.



                  When compiler optimizations are turned off, the F# compiler will produce different IL for each snippet. In other words, if there are any optimizations going on, it's left up to the JITter, and it may very well be slower to call the first form.



                  However, when compiler optimizations are turned on, both forms produce identical IL outputs in every scenario I could think of to test it. In fact, with both forms, calling:



                  add 3 4


                  yields the IL equivalent of a hard-coded 7, with the entire function call optimized away:



                  ldc.i4.7


                  In other words, the F# compiler is pretty thorough when it comes to optimizing logically identical code blocks.



                  This is not an exhaustive answer, of course, and there could be some case where they are actually treated differently by the compiler.






                  share|improve this answer












                  I tested this in LINQPad 5.



                  When compiler optimizations are turned off, the F# compiler will produce different IL for each snippet. In other words, if there are any optimizations going on, it's left up to the JITter, and it may very well be slower to call the first form.



                  However, when compiler optimizations are turned on, both forms produce identical IL outputs in every scenario I could think of to test it. In fact, with both forms, calling:



                  add 3 4


                  yields the IL equivalent of a hard-coded 7, with the entire function call optimized away:



                  ldc.i4.7


                  In other words, the F# compiler is pretty thorough when it comes to optimizing logically identical code blocks.



                  This is not an exhaustive answer, of course, and there could be some case where they are actually treated differently by the compiler.







                  share|improve this answer












                  share|improve this answer



                  share|improve this answer










                  answered Nov 28 at 20:57









                  p.s.w.g

                  114k19206252




                  114k19206252






























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