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Author Topic: [Tutorial] Edition 2: Using Axe to its full functional form λ  (Read 21149 times)

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Offline ztrumpet

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #30 on: July 28, 2011, 09:55:39 pm »
Quote from: Ashbad on July 28, 2011, 09:33:52 pm
So, is this just going to never be in the new "Tutorials" section?  I've asked like 5 times in 3 separate threads, and I'm pretty much forced to give up all hope.  People who want to see this will have to follow the new "this is outdated link" in the old tutorial, I guess :/
Nope, you can add it now. :D

If you'd like, I can add it for you.  Just let me know. :)
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Ashbad

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #31 on: July 28, 2011, 09:57:54 pm »
Quote from: ztrumpet on July 28, 2011, 09:55:39 pm
Quote from: Ashbad on July 28, 2011, 09:33:52 pm
So, is this just going to never be in the new "Tutorials" section?  I've asked like 5 times in 3 separate threads, and I'm pretty much forced to give up all hope.  People who want to see this will have to follow the new "this is outdated link" in the old tutorial, I guess :/
Nope, you can add it now. :D

If you'd like, I can add it for you.  Just let me know. :)

orly?

I didn't know I could add this.. Sorry for the bitching :P
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Offline ztrumpet

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #32 on: July 28, 2011, 10:06:17 pm »
Quote from: Ashbad on July 28, 2011, 09:57:54 pm
Quote from: ztrumpet on July 28, 2011, 09:55:39 pm
Quote from: Ashbad on July 28, 2011, 09:33:52 pm
So, is this just going to never be in the new "Tutorials" section?  I've asked like 5 times in 3 separate threads, and I'm pretty much forced to give up all hope.  People who want to see this will have to follow the new "this is outdated link" in the old tutorial, I guess :/
Nope, you can add it now. :D

If you'd like, I can add it for you.  Just let me know. :)

orly?

I didn't know I could add this.. Sorry for the bitching :P
It actually just got changed: http://ourl.ca/12282

I didn't know how to do it before, so that's why I never offered until now. ;)
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Offline DJ Omnimaga

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #33 on: July 28, 2011, 10:42:12 pm »
Well to be honest after I left the tutorials section kinda got abandonned. No more tutorials got added, but now that they can be added by their respective authors too, this should help a lot speeding up the process.
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Offline WaeV

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #34 on: September 19, 2011, 09:33:20 pm »
Hey guys, I was really impressed by these new features of Axe, so I decided to register to make this post!  I figured we could use a non-trivial example of how lambdas can be actually used.


This is an example of some functional code that we wrote in one of my programming classes.  The objective was to write a function that creates iterators.  Each time you call the iterator, it returns one more than the previous call.  We do this with lambda syntax and closures.  We wrote this in Scheme, but I was surprised and impressed to see it correctly execute in Axe.

The Axe code:
(I used liberal amounts of spaces to get it to line up the way I did)
Code: [Select]
.ITERATE

Iter()→L

Disp (L)()►Dec,i
Disp (L)()►Dec,i
Disp (L)()►Dec,i

Iter()→M

Disp (M)()►Dec,i
Disp (M)()►Dec,i
Disp (L)()►Dec,i

Lbl Iter
Return
λ(
  λ(
    λ(
      r₁+1→r₁
    )
  )(0)
)()
Output:
Code: [Select]
asm(prgmITERATE
    1
    2
    3
    1
    2
    4
An interesting side-effect of this is that the current iteration is not stored in any variables - it's captured in the closure.  Not that it's efficient to do this on a calculator with limited memory, but...
« Last Edit: September 20, 2011, 06:32:23 pm by WaeV »
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Offline Michael_Lee

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #35 on: September 20, 2011, 09:52:41 pm »
That's pretty cool!

You should post that in the Axe Routines Thread

I don't know much about functional programming -- is it possible to pass in an arbitary lambda to the iterator generator so that it can iterate over a sequence of numbers other then just '1 2 3 4'?

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Offline WaeV

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #36 on: September 21, 2011, 02:11:02 am »
Yeah sure!  You can do that. The outer lambda returns the inner lambda - your phrase "iterator generator" is a good one - but it could return any lambda.

Edit: Here's a version which lets you decide where to start.
The Axe code:
Code: [Select]
.ITERATE

Iter(10)→L

Disp (L)()►Dec,i
Disp (L)()►Dec,i
Disp (L)()►Dec,i

Lbl Iter
Return
λ(
  λ(
    λ(
      r₁+1→r₁
    )
  )(r₁)
)()
Output:
Code: [Select]
asm(prgmITERATE
    11
    12
    13
« Last Edit: September 21, 2011, 06:20:00 pm by WaeV »
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Offline epic7

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #37 on: October 23, 2011, 10:08:27 pm »
What are the r1, r2, etc.?
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Offline Deep Toaster

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #38 on: October 23, 2011, 11:02:35 pm »
Subroutine arguments, as in sub(LBL,1,2,3) calls LBL with 1 in r1, 2 in r2, and 3 in r3.
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Offline Builderboy

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #39 on: October 23, 2011, 11:27:52 pm »
Quote from: WaeV on September 19, 2011, 09:33:20 pm
Hey guys, I was really impressed by these new features of Axe, so I decided to register to make this post!  I figured we could use a non-trivial example of how lambdas can be actually used.


This is an example of some functional code that we wrote in one of my programming classes.  The objective was to write a function that creates iterators.  Each time you call the iterator, it returns one more than the previous call.  We do this with lambda syntax and closures.  We wrote this in Scheme, but I was surprised and impressed to see it correctly execute in Axe.

The Axe code:
(I used liberal amounts of spaces to get it to line up the way I did)
Code: [Select]
.ITERATE

Iter()→L

Disp (L)()►Dec,i
Disp (L)()►Dec,i
Disp (L)()►Dec,i

Iter()→M

Disp (M)()►Dec,i
Disp (M)()►Dec,i
Disp (L)()►Dec,i

Lbl Iter
Return
λ(
  λ(
    λ(
      r₁+1→r₁
    )
  )(0)
)()
Output:
Code: [Select]
asm(prgmITERATE
    1
    2
    3
    1
    2
    4
An interesting side-effect of this is that the current iteration is not stored in any variables - it's captured in the closure.  Not that it's efficient to do this on a calculator with limited memory, but...

I can't replicate your output?  I get 1,2,3,1,2,3.  I think the issue is that you believe r1-r6 to be local variables to each function, but in fact they are global?
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Offline epic7

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #40 on: October 24, 2011, 07:46:08 pm »
Quote from: Deep Thought on October 23, 2011, 11:02:35 pm
Subroutine arguments, as in sub(LBL,1,2,3) calls LBL with 1 in r1, 2 in r2, and 3 in r3.

What's a subrouine argument?P
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Offline aeTIos

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #41 on: October 25, 2011, 10:52:00 am »
a subroutine argument is a value you give to a subroutine.
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Offline LincolnB

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #42 on: October 25, 2011, 06:09:03 pm »
Do this:

prgmTEMP

.STUFF

ClrHome

Stuff(1)

Pause 3000
Return

Lbl Stuff
Disp r1>Dec
Return

..end of prgmTEMP

Compile it, run prgmSTUFF, and it will display the letter 1.
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Offline epic7

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #43 on: October 25, 2011, 06:51:32 pm »
Quote from: aeTIos on October 25, 2011, 10:52:00 am
a subroutine argument is a value you give to a subroutine.

How would you use a value given to a subroutine?
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Offline Builderboy

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Re: [Tutorial] Edition 2: Using Axe to its full functional form λ
« Reply #44 on: October 25, 2011, 07:14:43 pm »
epic7, have you read the Axe user guide?  It has a lot of answers to a lot of the questions you have been having :)  To pass an argument to a subroutine, you would type Sub(ARG) where ARG is put into r1 when the routine is called
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