Generate tail call opcode

Question

Out of curiosity I was trying to generate a tail call opcode using C#. Fibinacci is an easy one, so my c# example looks like this:

    private static void M         


        

Answer 1

C# compiler does not give you any guarantees about tail-call optimizations because C# programs usually use loops and so they do not rely on the tail-call optimizations. So, in C#, this is simply a JIT optimization that may or may not happen (and you cannot rely on it).

F# compiler is designed to handle functional code that uses recursion and so it does give you certain guarantees about tail-calls. This is done in two ways:

• if you write a recursive function that calls itself (like your fib) the compiler turns it into a function that uses loop in the body (this is a simple optimization and the produced code is faster than using a tail-call)

• if you use a recursive call in a more complex position (when using continuation passing style where function is passed as an argument), then the compiler generates a tail-call instruction that tells the JIT that it must use a tail call.

As an example of the second case, compile the following simple F# function (F# does not do this in Debug mode to simplify debugging, so you may need Release mode or add --tailcalls+):

let foo a cont = cont (a + 1)

The function simply calls the function cont with the first argument incremented by one. In continuation passing style, you have a long sequence of such calls, so the optimization is crucial (you simply cannot use this style without some handling of tail calls). The generates IL code looks like this:

IL_0000: ldarg.1
IL_0001: ldarg.0
IL_0002: ldc.i4.1
IL_0003: add
IL_0004: tail.                          // Here is the 'tail' opcode!
IL_0006: callvirt instance !1 
  class [FSharp.Core] Microsoft.FSharp.Core.FSharpFunc`2::Invoke(!0)
IL_000b: ret