What is "H" in opcodes (very confused) - opcode

In my computer organization class, we're learning opcodes, and i understand most of the opcodes, and the concepts of them, but seriously, theres no legend for what "H" is for example..
iadd
iadd1:mar=sp=sp-1;rd
iadd2: H =tos..
whats H??
x'D thanks!

H is a register name, not an opcode.
Hold (H)
This register holds data that is to be supplied to the left (A) input of the arithmetic logic unit. This is the only register that can perform this function.
http://dsearls.org/courses/C391OrgSys/SimHelp/HTML/IJVM_Register_Set.htm

Related

Elixir allow assign to variable two times [duplicate]

In Dave Thomas's book Programming Elixir he states "Elixir enforces immutable data" and goes on to say:
In Elixir, once a variable references a list such as [1,2,3], you know it will always reference those same values (until you rebind the variable).
This sounds like "it won't ever change unless you change it" so I'm confused as to what the difference between mutability and rebinding is. An example highlighting the differences would be really helpful.
Don't think of "variables" in Elixir as variables in imperative languages, "spaces for values". Rather look at them as "labels for values".
Maybe you would better understand it when you look at how variables ("labels") work in Erlang. Whenever you bind a "label" to a value, it remains bound to it forever (scope rules apply here of course).
In Erlang you cannot write this:
v = 1, % value "1" is now "labelled" "v"
% wherever you write "1", you can write "v" and vice versa
% the "label" and its value are interchangeable
v = v+1, % you can not change the label (rebind it)
v = v*10, % you can not change the label (rebind it)
instead you must write this:
v1 = 1, % value "1" is now labelled "v1"
v2 = v1+1, % value "2" is now labelled "v2"
v3 = v2*10, % value "20" is now labelled "v3"
As you can see this is very inconvenient, mainly for code refactoring. If you want to insert a new line after the first line, you would have to renumber all the v* or write something like "v1a = ..."
So in Elixir you can rebind variables (change the meaning of the "label"), mainly for your convenience:
v = 1 # value "1" is now labelled "v"
v = v+1 # label "v" is changed: now "2" is labelled "v"
v = v*10 # value "20" is now labelled "v"
Summary: In imperative languages, variables are like named suitcases: you have a suitcase named "v". At first you put sandwich in it. Than you put an apple in it (the sandwich is lost and perhaps eaten by the garbage collector). In Erlang and Elixir, the variable is not a place to put something in. It's just a name/label for a value. In Elixir you can change a meaning of the label. In Erlang you cannot. That's the reason why it doesn't make sense to "allocate memory for a variable" in either Erlang or Elixir, because variables do not occupy space. Values do. Now perhaps you see the difference clearly.
If you want to dig deeper:
1) Look at how "unbound" and "bound" variables work in Prolog. This is the source of this maybe slightly strange Erlang concept of "variables which do not vary".
2) Note that "=" in Erlang really is not an assignment operator, it's just a match operator! When matching an unbound variable with a value, you bind the variable to that value. Matching a bound variable is just like matching a value it's bound to. So this will yield a match error:
v = 1,
v = 2, % in fact this is matching: 1 = 2
3) It's not the case in Elixir. So in Elixir there must be a special syntax to force matching:
v = 1
v = 2 # rebinding variable to 2
^v = 3 # matching: 2 = 3 -> error
Immutability means that data structures don't change. For example the function HashSet.new returns an empty set and as long as you hold on to the reference to that set it will never become non-empty. What you can do in Elixir though is to throw away a variable reference to something and rebind it to a new reference. For example:
s = HashSet.new
s = HashSet.put(s, :element)
s # => #HashSet<[:element]>
What cannot happen is the value under that reference changing without you explicitly rebinding it:
s = HashSet.new
ImpossibleModule.impossible_function(s)
s # => #HashSet<[:element]> will never be returned, instead you always get #HashSet<[]>
Contrast this with Ruby, where you can do something like the following:
s = Set.new
s.add(:element)
s # => #<Set: {:element}>
Erlang and obviously Elixir that is built on top of it, embraces immutability.
They simply don’t allow values in a certain memory location to change. Never Until the variable gets garbage collected or is out of scope.
Variables aren't the immutable thing. The data they point to is the immutable thing. That's why changing a variable is referred to as rebinding.
You're point it at something else, not changing the thing it points to.
x = 1 followed by x = 2 doesn't change the data stored in computer memory where the 1 was to a 2. It puts a 2 in a new place and points x at it.
x is only accessible by one process at a time so this has no impact on concurrency and concurrency is the main place to even care if something is immutable anyway.
Rebinding doesn’t change the state of an object at all, the value is still in the same memory location, but it’s label (variable) now points to another memory location, so immutability is preserved. Rebinding is not available in Erlang, but while it is in Elixir this is not braking any constraint imposed by the Erlang VM, thanks to its implementation.
The reasons behind this choice are well explained by Josè Valim in this gist .
Let's say you had a list
l = [1, 2, 3]
and you had another process that was taking lists and then performing "stuff" against them repeatedly and changing them during this process would be bad. You might send that list like
send(worker, {:dostuff, l})
Now, your next bit of code might want to update l with more values for further work that's unrelated to what that other process is doing.
l = l ++ [4, 5, 6]
Oh no, now that first process is going to have undefined behavior because you changed the list right? Wrong.
That original list remains unchanged. What you really did was make a new list based on the old one and rebind l to that new list.
The separate process never has access to l. The data l originally pointed at is unchanged and the other process (presumably, unless it ignored it) has its own separate reference to that original list.
What matters is you can't share data across processes and then change it while another process is looking at it. In a language like Java where you have some mutable types (all primitive types plus references themselves) it would be possible to share a structure/object that contained say an int and change that int from one thread while another was reading it.
In fact, it's possible to change a large integer type in java partially while it's read by another thread. Or at least, it used to be, not sure if they clamped that aspect of things down with the 64 bit transition. Anyway, point is, you can pull the rug out from under other processes/threads by changing data in a place that both are looking at simultaneously.
That's not possible in Erlang and by extension Elixir. That's what immutability means here.
To be a bit more specific, in Erlang (the original language for the VM Elixir runs on) everything was single-assignment immutable variables and Elixir is hiding a pattern Erlang programmers developed to work around this.
In Erlang, if a=3 then that was what a was going to be its value for the duration of that variable's existence until it dropped out of scope and was garbage collected.
This was useful at times (nothing changes after assignment or pattern match so it is easy to reason about what a function is doing) but also a bit cumbersome if you were doing multiple things to a variable or collection over the course executing a function.
Code would often look like this:
A=input,
A1=do_something(A),
A2=do_something_else(A1),
A3=more_of_the_same(A2)
This was a bit clunky and made refactoring more difficult than it needed to be. Elixir is doing this behind the scenes, but hiding it from the programmer via macros and code transforms performed by the compiler.
Great discussion here
immutability-in-elixir
The variables really are immutable in sense, every new rebinding (assignment) is only visible to access that come after that. All previous access, still refer to old value(s) at the time of their call.
foo = 1
call_1 = fn -> IO.puts(foo) end
foo = 2
call_2 = fn -> IO.puts(foo) end
foo = 3
foo = foo + 1
call_3 = fn -> IO.puts(foo) end
call_1.() #prints 1
call_2.() #prints 2
call_3.() #prints 4
To make it a very simple
variables in elixir are not like container where you keep adding and removing or modifying items from the container.
Instead they are like Labels attached to a container, when you reassign a variable is as simple a you pick a label from one container and place it on a new container with expected data in it.

In GHC's STG output with -O2, what's this sequence following Str=DmdType all about?

(Misleading title: it's only one of a plethora of inter-related similar questions below: these sound like asking for a full reference manual but keep in mind for this topic there is no reference manual other than the entirety of GHC's source-codes of its STG pipeline stage, and the collective accumulated experience of others/"insiders"..)
I'm exploring "transpiling" Haskell (from scratch for fun/learning, ignoring existing projects; target language/s similarly high-level / "already-fit-for-STG-machine" with existing GC + lambdas/func-values + closures) and so I'm trying to become ever more familiar with GHC's STG IR. Having repeatedly gone through the dozen-or-two online articles/videos of varying age, depth, detail that actually deal with the topic (plus the original paper, plus StgSyn.hs), and understanding many-perhaps-most basic principles, seeing -ddump-stged output still baffles me in various parts (I won't manually parse it but reuse GHC API's in-memory AST later on of course) --- mostly I think I'm stuck mapping my "roughly known" concepts to the "still-foreign" abbreviated/codified identifiers of that IR. If you know your way around STG a bit, mind looking at the following mini-sample to clarify a few open questions and help further solidify my (and future searchers') grasp?
From a most simple .hs module, I have -ddump-stged twice, first (on the left) with -O0 and then (on the right) with -O2, both captured in this diff.
Walking through everything def-by-def..
Lines L_|R5-11: so in O2, testX1 and testX2 seem to be global constants/literals for the integers 4 and 5 --- O0 doesn't have them. Curious!
Is Str=DmdType something about strictness? "Strictness is of type on-demand" or some such? But then a top-level/heap-ish/"global" constant literal can't be "lazy" can it.. (one of the things where I can't just casually Ctrl+F in StgSyn.hs --- it's not in there! which is odd in its own way, how come there's STG syntax not in StgSyn.hs)
Caf have a rough idea about constant-applicative-forms, but Unf=OtherCon? "Other constructor" (unboxed/native Type.S#-related?) ..
Line L6|R14: Surprised to still see type-class information in there (Num), is that "just info/annotation" or is this crucial for any of the built-in code-gens to set up some "dictionary" lookup machinery at runtime? (I'd sure hope by the late STG / pre-CMM stage that would be resolved and inlined already where possible at least in O2. After all GHC has also decided to type-default 4 and 5 to Integer). Generally speaking I understand STG is "untyped" other than denoting prim types, saturated cons, perhaps strings (looks like it later on at the bottom), so such "typeclass" annotations can only be.. I guess for readers to find their way around the ddump-ed *.stg. But correct me if not.
GblId probably just "global identifier" aka top-level CAF right? Arity clear.
Line L7|R18: now Str=DmdType for testX is, only in O2, followed by a freakish <S(LLC(C(S))LLLL),U(1*C1(C1(U)),A,1*C1(C1(U)),A,A,A,C(U))><L,U>! What's that, SKI calculus? ;D no seriously, LLC.. LLLL.. stack or other memory layout hints for CMM? Any idea? Must be some optimization, would like to understand which-and-how..
Line L8|R20: $dNum_sGM (left) and $dNum_sIx (right) have me a bit concerned, they don't seem to be "defined at the module level" here anywhere. Typeclass "method dispatch dictionary lookup" kind of thing? Would eg. CMM take this together with the above Num annotation to set things up? It always appears together with the input func arg.
The whole function "body" for both left and right can be seen here essentially as "3 lets with a lambda-ish form for 3 atoms, 2 of which are statically known literal-constants" --- I suppose this is standard and to be expected in the STG IR AST? For the first of these, funnily enough we could say that O0 has "inlined the global (what is testX1 or testX2 in O2) and O2 hasn't" (making the latter much shorter as that applies to both these constant literals).
I've only ever seen Occ=Once, what are the others and how to interpret? Once for one isn't even in StgSyn.hs..
Now LclId a counterpart to the earlier encountered GblId. That's denoting the scope of the identifier? Could it also be anything else, in this expression context? As in: if traversing the AST I roughly know how deep I am, I can ignore this since if I'm at the top-level it must be GblId and otherwise LclId? Hm.. maybe better take what STG gives me but then I need to be sure about the semantics and possibilities.. guys, using StgSyn.hs I have the wrong source file, right? Nothing on this in there either.. (always hopeful as its comments are quite well-done)
the rest is just metadata as string constants, OK.. oh wait, look at O2, there's Str=DmdType m1 and Str=DmdType m, what's the m/m1 about, another thing I don't see "defined anywhere at the module level" here? And it's not in O0..
still going strong? Merely a bonus question (for now), tell us about srt:SRT:[] ;)
Just a few tidbits - a full answer is quite beyond my knowledge.
The type of your function is
testX :: GHC.Num.Num a => a -> a
It’s compiled to a function with two arguments: a dictionary of the Num type class, and the actual argument.
The $d… names stand for dictionaries of type class instances. The <S(LLC(C(S))LLLL),… annotations are strictness information about the function arguments. They basically say which part of the argument will be used by your function and which not. Looks a bit weird here because it contains information about all the class instance members.
Some of this is explained here:
https://ghc.haskell.org/trac/ghc/wiki/Commentary/Compiler/Demand
The str:STR: is the „Static reference table“, i.e. list of free variables of the expression - in your case, always [].

Prolog Accumulators. Are they really a "different" concept?

I am learning Prolog under my Artificial Intelligence Lab, from the source Learn Prolog Now!.
In the 5th Chapter we come to learn about Accumulators. And as an example, these two code snippets are given.
To Find the Length of a List
without accumulators:
len([],0).
len([_|T],N) :- len(T,X), N is X+1.
with accumulators:
accLen([_|T],A,L) :- Anew is A+1, accLen(T,Anew,L).
accLen([],A,A).
I am unable to understand, how the two snippets are conceptually different? What exactly an accumulator is doing different? And what are the benefits?
Accumulators sound like intermediate variables. (Correct me if I am wrong.) And I had already used them in my programs up till now, so is it really that big a concept?
When you give something a name, it suddenly becomes more real than it used to be. Discussing something can now be done by simply using the name of the concept. Without getting any more philosophical, no, there is nothing special about accumulators, but they are useful.
In practice, going through a list without an accumulator:
foo([]).
foo([H|T]) :-
foo(T).
The head of the list is left behind, and cannot be accessed by the recursive call. At each level of recursion you only see what is left of the list.
Using an accumulator:
bar([], _Acc).
bar([H|T], Acc) :-
bar(T, [H|Acc]).
At every recursive step, you have the remaining list and all the elements you have gone through. In your len/3 example, you only keep the count, not the actual elements, as this is all you need.
Some predicates (like len/3) can be made tail-recursive with accumulators: you don't need to wait for the end of your input (exhaust all elements of the list) to do the actual work, instead doing it incrementally as you get the input. Prolog doesn't have to leave values on the stack and can do tail-call optimization for you.
Search algorithms that need to know the "path so far" (or any algorithm that needs to have a state) use a more general form of the same technique, by providing an "intermediate result" to the recursive call. A run-length encoder, for example, could be defined as:
rle([], []).
rle([First|Rest],Encoded):-
rle_1(Rest, First, 1, Encoded).
rle_1([], Last, N, [Last-N]).
rle_1([H|T], Prev, N, Encoded) :-
( dif(H, Prev)
-> Encoded = [Prev-N|Rest],
rle_1(T, H, 1, Rest)
; succ(N, N1),
rle_1(T, H, N1, Encoded)
).
Hope that helps.
TL;DR: yes, they are.
Imagine you are to go from a city A on the left to a city B on the right, and you want to know the distance between the two in advance. How are you to achieve this?
A mathematician in such a position employs magic known as structural recursion. He says to himself, what if I'll send my own copy one step closer towards the city B, and ask it of its distance to the city? I will then add 1 to its result, after receiving it from my copy, since I have sent it one step closer towards the city, and will know my answer without having moved an inch! Of course if I am already at the city gates, I won't send any copies of me anywhere since I'll know that the distance is 0 - without having moved an inch!
And how do I know that my copy-of-me will succeed? Simply because he will follow the same exact rules, while starting from a point closer to our destination. Whatever value my answer will be, his will be one less, and only a finite number of copies of us will be called into action - because the distance between the cities is finite. So the total operation is certain to complete in a finite amount of time and I will get my answer. Because getting your answer after an infinite time has passed, is not getting it at all - ever.
And now, having found out his answer in advance, our cautious magician mathematician is ready to embark on his safe (now!) journey.
But that of course wasn't magic at all - it's all being a dirty trick! He didn't find out the answer in advance out of thin air - he has sent out the whole stack of others to find it for him. The grueling work had to be done after all, he just pretended not to be aware of it. The distance was traveled. Moreover, the distance back had to be traveled too, for each copy to tell their result to their master, the result being actually created on the way back from the destination. All this before our fake magician had ever started walking himself. How's that for a team effort. For him it could seem like a sweet deal. But overall...
So that's how the magician mathematician thinks. But his dual the brave traveler just goes on a journey, and counts his steps along the way, adding 1 to the current steps counter on each step, before the rest of his actual journey. There's no pretense anymore. The journey may be finite, or it may be infinite - he has no way of knowing upfront. But at each point along his route, and hence when ⁄ if he arrives at the city B gates too, he will know his distance traveled so far. And he certainly won't have to go back all the way to the beginning of the road to tell himself the result.
And that's the difference between the structural recursion of the first, and tail recursion with accumulator ⁄ tail recursion modulo cons ⁄ corecursion employed by the second. The knowledge of the first is built on the way back from the goal; of the second - on the way forth from the starting point, towards the goal. The journey is the destination.
see also:
Technical Report TR19: Unwinding Structured Recursions into Iterations. Daniel P. Friedman and David S. Wise (Dec 1974).
What are the practical implications of all this, you ask? Why, imagine our friend the magician mathematician needs to boil some eggs. He has a pot; a faucet; a hot plate; and some eggs. What is he to do?
Well, it's easy - he'll just put eggs into the pot, add some water from the faucet into it and will put it on the hot plate.
And what if he's already given a pot with eggs and water in it? Why, it's even easier to him - he'll just take the eggs out, pour out the water, and will end up with the problem he already knows how to solve! Pure magic, isn't it!
Before we laugh at the poor chap, we mustn't forget the tale of the centipede. Sometimes ignorance is bliss. But when the required knowledge is simple and "one-dimensional" like the distance here, it'd be a crime to pretend to have no memory at all.
accumulators are intermediate variables, and are an important (read basic) topic in Prolog because allow reversing the information flow of some fundamental algorithm, with important consequences for the efficiency of the program.
Take reversing a list as example
nrev([],[]).
nrev([H|T], R) :- nrev(T, S), append(S, [H], R).
rev(L, R) :- rev(L, [], R).
rev([], R, R).
rev([H|T], C, R) :- rev(T, [H|C], R).
nrev/2 (naive reverse) it's O(N^2), where N is list length, while rev/2 it's O(N).

understanding referential transparency

Generally, I have a headache because something is wrong with my reasoning:
For 1 set of arguments, referential transparent function will always return 1 set of output values.
that means that such function could be represented as a truth table (a table where 1 set of output parameters is specified for 1 set of arguments).
that makes the logic behind such functions is combinational (as opposed to sequential)
that means that with pure functional language (that has only rt functions) it is possible to describe only combinational logic.
The last statement is derived from this reasoning, but it's obviously false; that means there is an error in reasoning. [question: where is error in this reasoning?]
UPD2. You, guys, are saying lots of interesting stuff, but not answering my question. I defined it more explicitly now. Sorry for messing up with question definition!
Question: where is error in this reasoning?
A referentially transparent function might require an infinite truth table to represent its behavior. You will be hard pressed to design an infinite circuit in combinatory logic.
Another error: the behavior of sequential logic can be represented purely functionally as a function from states to states. The fact that in the implementation these states occur sequentially in time does not prevent one from defining a purely referentially transparent function which describes how state evolves over time.
Edit: Although I apparently missed the bullseye on the actual question, I think my answer is pretty good, so I'm keeping it :-) (see below).
I guess a more concise way to phrase the question might be: can a purely functional language compute anything an imperative one can?
First of all, suppose you took an imperative language like C and made it so you can't alter variables after defining them. E.g.:
int i;
for (i = 0; // okay, that's one assignment
i < 10; // just looking, that's all
i++) // BUZZZ! Sorry, can't do that!
Well, there goes your for loop. Do we get to keep our while loop?
while (i < 10)
Sure, but it's not very useful. i can't change, so it's either going to run forever or not run at all.
How about recursion? Yes, you get to keep recursion, and it's still plenty useful:
int sum(int *items, unsigned int count)
{
if (count) {
// count the first item and sum the rest
return *items + sum(items + 1, count - 1);
} else {
// no items
return 0;
}
}
Now, with functions, we don't alter state, but variables can, well, vary. Once a variable passes into our function, it's locked in. However, we can call the function again (recursion), and it's like getting a brand new set of variables (the old ones stay the same). Although there are multiple instances of items and count, sum((int[]){1,2,3}, 3) will always evaluate to 6, so you can replace that expression with 6 if you like.
Can we still do anything we want? I'm not 100% sure, but I think the answer is "yes". You certainly can if you have closures, though.
You have it right. The idea is, once a variable is defined, it can't be redefined. A referentially transparent expression, given the same variables, always yields the same result value.
I recommend looking into Haskell, a purely functional language. Haskell doesn't have an "assignment" operator, strictly speaking. For instance:
my_sum numbers = ??? where
i = 0
total = 0
Here, you can't write a "for loop" that increments i and total as it goes along. All is not lost, though. Just use recursion to keep getting new is and totals:
my_sum numbers = f 0 0 where
f i total =
if i < length numbers
then f i' total'
else total
where
i' = i+1
total' = total + (numbers !! i)
(Note that this is a stupid way to sum a list in Haskell, but it demonstrates a method of coping with single assignment.)
Now, consider this highly imperative-looking code:
main = do
a <- readLn
b <- readLn
print (a + b)
It's actually syntactic sugar for:
main =
readLn >>= (\a ->
readLn >>= (\b ->
print (a + b)))
The idea is, instead of main being a function consisting of a list of statements, main is an IO action that Haskell executes, and actions are defined and chained together with bind operations. Also, an action that does nothing, yielding an arbitrary value, can be defined with the return function.
Note that bind and return aren't specific to actions. They can be used with any type that calls itself a Monad to do all sorts of funky things.
To clarify, consider readLn. readLn is an action that, if executed, would read a line from standard input and yield its parsed value. To do something with that value, we can't store it in a variable because that would violate referential transparency:
a = readLn
If this were allowed, a's value would depend on the world and would be different every time we called readLn, meaning readLn wouldn't be referentially transparent.
Instead, we bind the readLn action to a function that deals with the action, yielding a new action, like so:
readLn >>= (\x -> print (x + 1))
The result of this expression is an action value. If Haskell got off the couch and performed this action, it would read an integer, increment it, and print it. By binding the result of an action to a function that does something with the result, we get to keep referential transparency while playing around in the world of state.
As far as I understand it, referential transparency just means: A given function will always yield the same result when invoked with the same arguments. So, the mathematical functions you learned about in school are referentially transparent.
A language you could check out in order to learn how things are done in a purely functional language would be Haskell. There are ways to use "updateable storage possibilities" like the Reader Monad, and the State Monad for example. If you're interested in purely functional data structures, Okasaki might be a good read.
And yes, you're right: Order of evaluation in a purely functional language like haskell does not matter as in non-functional languages, because if there are no side effects, there is no reason to do someting before/after something else -- unless the input of one depends on the output of the other, or means like monads come into play.
I don't really know about the truth-table question.
Here's my stab at answering the question:
Any system can be described as a combinatorial function, large or small.
There's nothing wrong with the reasoning that pure functions can only deal with combinatorial logic -- it's true, just that functional languages hide that from you to some extent or another.
You could even describe, say, the workings of a game engine as a truth table or a combinatorial function.
You might have a deterministic function that takes in "the current state of the entire game" as the RAM occupied by the game engine and the keyboard input, and returns "the state of the game one frame later". The return value would be determined by the combinations of the bits in the input.
Of course, in any meaningful and sane function, the input is parsed down to blocks of integers, decimals and booleans, but the combinations of the bits in those values is still determining the output of your function.
Keep in mind also that basic digital logic can be described in truth tables. The only reason that that's not done for anything more than, say, arithmetic on 4-bit integers, is because the size of the truth table grows exponentially.
The error in Your reasoning is the following:
"that means that such function could be represented as a truth table".
You conclude that from a functional language's property of referential transparency. So far the conclusion would sound plausible, but You oversee that a function is able to accept collections as input and process them in contrast to the fixed inputs of a logic gate.
Therefore a function does not equal a logic gate but rather a construction plan of such a logic gate depending on the actual (at runtime determined) input!
To comment on Your comment: Functional languages can - although stateless - implement a state machine by constructing the states from scratch each time they are being accessed.

What, if any, is wrong with this approach to declarative I/O

I'm not sure exactly how much this falls under 'programming' opposed to 'program language design'. But the issue is this:
Say, for sake of simplicity we have two 'special' lists/arrays/vectors/whatever we just call 'ports' for simplicity, one called stdIn and another stdOut. These conceptually represent respectively
All the user input given to the program in the duration of the program
All the output written to the terminal during the duration of the program
In Haskell-inspired pseudocode, it should then be possible to create this wholly declarative program:
let stdOut = ["please input a number",
"and please input another number",
"The product of both numbers is: " ++ stdIn[0] * stdIn[1]]
Which would do the expected, ask for two numbers, and print their product. The trick being that stdOut represents the list of strings written to the terminal at the completion of the program, and stdIn the list of input strings. Type errors and the fact that there needs to be some safeguard to only print the next line after a new line has been entered left aside here for simplicity's sake, it's probably easy enough to solve that.
So, before I go of to implement this idea, are there any pitfalls to it that I overlooked? I'm not aware of a similar construct already existing so it'd be naïve to not take into account that there is an obvious pitfall to it I overlooked.
Otherwise, I know that of course:
let stdOut = [stdIn[50],"Hello, World!"]
Would be an error if these results need to be interwoven in a similar fashion as above.
A similar approach was used in early versions of Haskell, except that the elements of the stdin and stdout channels were not strings but generic IO 'actions'--in fact, input and output were generalized to 'response' and 'request'. As long as both channels are lazy (i.e. they are actually 'iterators' or 'enumerators'), the runtime can simply walk the request channel, act on each request and tack appropriate responses onto the response channel. Unfortunately, the system was very hard to use, so it was scrapped in favor of monadic IO. See these papers:
Hudak, P., and Sundaresh, R. On the expressiveness of purely-functional I/O systems. Tech. Rep. YALEU/DCS/RR-665, Department of Computer Science, Yale University, Mar. 1989.
Peyton Jones, S. Tackling the Awkward Squad: monadic input/output, concurrency, exceptions, and foreign-language calls in Haskell. In Engineering theories of software construction, 2002, pp. 47--96.
The approach you're describing sounds like "Dialogs." In their award-winning 1993 paper Imperative Functional Programming, Phil Wadler and Simon Peyton Jones give some examples where dialogs really don't work very well, and they explain why monadic I/O is better.
I don't see how you will weave them considering this example compared to your own:
let stdOut = ["Welcome to the program which multiplies.",
"please input a number",
"and please input another number",
"The product of both numbers is: " ++ stdIn[0] * stdIn[1]]
Should the program prompt for the number represented by stdIn[0] after outputting one line (as in your example) or two lines? If the index 0 represents the 0th input from stdin, then it seems something similar to:
let stdOut = ["Welcome to the program which multiplies.",
"please input a number",
some_annotation(stdIn[0]),
"and please input another number",
some_annotation(stdIn[1]),
"The product of both numbers is: " ++ stdIn[0] * stdIn[1]]
will be required in order to coordinate the timing of output and input.
I like your idea. Replace some_annotation with your preference, perhaps something akin "synchronize?" I couldn't come up with the incisive word for it.
This approach seems to be the "most obvious" way to add I/O to a pure λ-calculus, and other people have mentioned that something along those lines has been tried in Haskell and Miranda.
However, I am aware of a language, not based on a λ-calculus, that still uses a very similar system:
How to handle input and output in a
language without side effects? In a
certain sense, input and output aren't
side effects; they are, so to speak,
front- and back-effects. (...) [A program is]
a function from the space
of possible inputs to the space of
possible outputs.
Input and output streams are
represented as lists of natural
numbers from 0 to 255, each
corresponding to one byte. End-of-file
is represented by the value 256, not
by end of list. (This is because it is
often easier to deal with EOF as a
character than as a special case.
Nevertheless, I wonder if it wouldn't
be better to use end-of-list.)
(...)
It's not difficult to write
interactive programs (...) [but] doing
so is, technically speaking, a sin.
(...) In a referentially transparent
language, anything not explicitly
synchronized is fair game for
evaluation in any order whatsoever, at
the run-time system's discretion.
(...) The most obvious way of writing
this particular program is to cons
together the "Hello, [name]!" string
in an expression which is conditioned
on receipt of a newline. If you do
this you are safe, because there's no
way for any evaluator to prove in
advance that the user will ever type a
newline.
(...)
So there's no practical problem with
interactive software. Nevertheless,
there's something unpleasant about the
way the second case is prevented. A
referentially transparent program
should not have to rely on lazy
evaluation in order to work properly.
How to escape this moral dilemma? The
hard way is to switch to a more
sophisticated I/O system, perhaps
based on Haskell's, in which input and
output are explicitly synchronized.
I'm rather disinclined to do this, as
I much prefer the simplicity of the
current system. The easy way out is to
write batch programs which happen to
work well interactively. This is
mainly just a matter of not prompting
the user.
Perhaps you would enjoying doing some programming in Lazy K?

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