How to pack / unpack a hex string (high nibble first) with Elixir - hex

I was wondering how I would work with hex strings in Elixir. Specifically, I'm interested in converting from Hex to ASCII.
In Ruby, an implementation of this may be:
["001C7F616A8B002128C1A33E8100"].pack('H*').gsub(/[^[:print:]]/, '.')
How would I accomplish this task with Elixir? I have tried:
<<00, 01, C7, F6...>>
but this isn't a correct representation of the hex for a string. Thanks for your time and assistance!
So I've made some progress but am currently struggling with the recursion aspect of this.
This is my solution thus far:
defmodule ElixirNetworkTools do
def decode(payload) do
upper_payload = String.upcase payload
case Base.decode16(upper_payload) do
:error -> decode_with_nonprintable_characters(payload)
{:ok, decoded_payload} -> decoded_payload
end
|> IO.write
end
def decode_with_nonprintable_characters(payload) do
String.chunk(payload, ~r/\w{2}/)
|> Enum.each(fn(byte) ->
case Base.decode16(byte) do
:error -> '.'
{:ok, decoded_payload} -> decoded_payload
end
end)
end
end

Here is another solution to the problem. A couple things before we start:
You can pass case: :mixed to Base.decode16/2: Base.decode16(string, case: :mixed), for this reason, you don't need do upcase before.
If you are going to raise on an invalid string, don't bother checking, just call decode16 directly as it also checks the size.
This means we can start with:
decoded = Base.decode16!(string, case: :mixed)
Now you need to replace non-printable characters. Don't use String.printable?/1 because it is about UTF-8 and not ASCII. We need to implement our own function but what makes more sense: to raise or replace them? It seems it must be considered an error if someone send invalidate data? If that is the case:
def validate_ascii!(<<h, t::binary>>) when h <= 127 do
validate_ascii!(t)
end
def validate_ascii!(<<>>) do
true
end
def validate_ascii!(rest) do
raise "invalid ascii on string starting at: #{rest}"
end
Alternatively you can just remove the last clause and it fail too.
Now we can put it together:
decoded = Base.decode16!(string, case: :mixed)
validate_ascii!(decoded)
decoded
EDIT: If you need to replace non-ascii by dots:
def keep_ascii(<<h, t::binary>>, acc) when h <= 127 do
keep_ascii(t, acc <> <<h>>)
end
def keep_ascii(<<_, t::binary>>, acc) do
keep_ascii(t, acc <> ".")
end
def keep_ascii(<<>>, acc) do
acc
end

The solution ended up being as follows, though if there is a cleaner or better solution I would be very interested in knowing it.
defmodule ElixirNetworkTools do
#doc """
The decode function takes a hexadecimal payload, such as one generated
by Snort, and returns the ASCII representation of the string.
## Example
iex> ElixirNetworkTools.decode("436F6E74656E742D4C656E6774683A203132")
{:ok, "Content-Length: 12"}
"""
def decode(payload) do
case _validate_length_of_snort(payload) do
:error -> raise "Invalid length hex string. Must be even length. Exiting"
_ -> nil
end
decoded = String.upcase(payload)
|> _do_decode
|> to_string
{:ok, decoded}
end
#doc """
Internal function used to manually process the hexadecimal payload,
and builds a char list of the printable characters. If a character is
not printable, we instead use periods.
## Example
iex> ElixirNetworkTools._do_decode("436F6E74656E742D4C656E6774683A203132")
["Content-Length: 12"]
"""
def _do_decode(payload) do
Base.decode16!(payload)
|> String.chunk(:printable)
|> Enum.map(fn(chunk) ->
case String.printable? chunk do
true -> chunk
_ -> "."
end
end)
end
#doc """
Internal function used to validate the length of the hexadecimal payload.
Hexadecimal strings should have an even number of characters.
## Example
iex> ElixirNetworkTools._validate_length_of_snort("436F6E74656E742D4C656E6774683A203132")
:ok
"""
def _validate_length_of_snort(payload) do
String.length(payload)
|> rem(2)
|> case do
0 -> :ok
_ -> :error
end
end
end

Related

Returning values from elixir functions inside modules

This is the method that processes an input string
def process(input) do
list=String.split(input, "\n")
f3=fn(a) ->
String.split(a," ")
end
list=Enum.map(list, f3)
func3=fn(n) ->
length(n)==3
end
func2=fn(n) ->
length(n)<=2
end
system=for x <-list, func3.(x), do: x
input=for y <- list, func2.(y), do: y
input=Enum.slice(input,0..length(input)-2)
output=""
output(input,output, system)
end
This is the output function that uses recursion to edit a string and eventually return its value
def output(input, output, system) do
cond do
length(input)==0 ->
output
true ->
[thing|tail]=input
if length(thing)==2 do
output=output<>"From "<>Enum.at(thing, 0)<>" to "<>Enum.at(thing,1)<>" is "<>Integer.to_string(calculate(thing, system))<>"km\n"
output(tail, output, system)
end
if length(thing)==1 do
if Enum.at(thing,0)=="Sun" do
output=output<>"Sun orbits"
output(tail, output, system)
else
output=output<>orbits(thing, system)<>" Sun"
output(tail, output, system)
end
end
output(tail, output, system)
end
end
As you can see when the input is an empty list it should return the output string. Using inspect shows that the output string does indeed have the correct value. Yet when the function is called in process(), it only returns the empty string, or nil.
Any help is appreciated, I am new to elixir so apologies if my code is a bit messy.
This could be a case where using pattern matching in the function head will let you avoid essentially all of the conditionals. You could break this down as:
def output([], message, _) do
message
end
def output([[from, to] | tail], message, system) do
distance = Integer.to_string(calculate(thing, system))
new_message = "#{message}From #{from} to #{to} is #{distance} km\n"
output(tail, new_message, system)
end
def output([["Sun"] | tail], message, system) do
output(tail, "Sun orbits #{message}", system)
end
def output([[thing] | tail], message, system) do
new_message = "#{message}#{orbits([thing], system)} Sun"
output(tail, new_message, system)
end
This gets around some of the difficulties highlighted in the comments: reassigning output inside a block doesn't have an effect, and there aren't non-local returns so after an if ... end block completes and goes on to the next conditional, its result is lost. This will also trap some incorrect inputs, and your process will exit with a pattern-match error if an empty or 3-element list winds up in the input list.
I've renamed the output parameter to the output function to message. This isn't required – the code will work fine whether or not you change it – but I found it a little confusing reading through the function whether output is a function call or a variable.

Infinite loop with counter in elixir

I'm studying functional programming and I want to implement something like this.
while(true) do
if(somethingHappensHere) {
break
}
counter++
end
return counter
How can I do this in functional way using elixir?
Thanks for this.
While in most functional programming languages one would use a recursion for this task, Elixir particularly provides the way to do this without using an explicit recursion call: Enum.reduce_while/3:
Enum.reduce_while(1..100, 0, fn i, acc ->
if condition, do: {:halt, acc}, else: {:cont, acc + i}
end)
For lazy evaluation one would use Stream.reduce_while/3.
To make it infinite, one might use one of infinite generators, provided by Stream module, like Stream.iterate/2:
Stream.iterate(0, &(&1+1)) |> Enum.reduce_while(0, fn i, acc ->
if i > 6, do: {:halt, acc}, else: {:cont, acc + 1}
end)
#⇒ 7
For the sake of recursion, this is how the recursive solution might be implemented in Elixir:
defmodule M do
def checker, do: & &1 <= 0
def factorial(v, acc \\ 1) do
if checker().(v), do: acc, else: factorial(v - 1, v * acc)
end
end
M.factorial 6
#⇒ 720
Not sure about elixir specifically, but you can achieve this using recursion:
function myFunction(int counter)
{
if (condition) {
return counter
}
return myFunction(counter + 1)
}
This essentially sets up a function that can infinitely recurse (call itself), each time passing in the next counter value.
By having the recursive call as the last thing the function does, this is known as tail-call recursion which elixir supports (as per: Does Elixir infinite recursion ever overflow the stack?)
This can then be used as such:
int counterValue = myFunction(0)
With the function only returning once the condition is true.
You could also make this more generic by having the function take another function that returns true or false (i.e. performs the conditional check).
As I said, unfortunately I'm not aware of the syntax of elixir, but I'm sure you'll be able to bridge that gap.
An example of something in Elixir syntax:
defmodule SOQuestion do
def test(counter) do
if (something_happens_here?()), do: counter, else: test(counter+1)
end
def something_happens_here?() do
true
end
end
And it would be invoked like this:
SOQuestion.test(0)
A couple of notes on this:
1.) It's a code fragment. Obviously it's tough to be very complete given the broad nature of your question.
2.) something_happens_here? being a predicate it would normally be named ending with a question mark.
3.) If something_happens_here? were defined in a different module, then the call would be if (Module.something_happens_here?())
4.) I've obviously coded something_happens_here? to simply return true unconditionally. In real code, of course, you'd want to pass some argument to something_happens_here? and act on that to determine which boolean to return.
Given all that I totally agree with #mudasowba--this sort of thing is usually better handled with one of the higher order functions built into the language. It's less error prone and often much easier for others to read too.
As mentioned, you could use a number of built-in functions like Enum.reduce_while/3. However, sometimes it is just as easy (or fun) to use simple recursion.
Using recursion:
I will make some generic examples and use bar(foo) as an example of your somethingHappensHere condition.
1) If bar(foo) is something allowed in a guard clause:
defmodule Counter do
def count do
count(foo, 0)
end
defp count(foo, count) when bar(foo), do: count
defp count(foo, count), do: count(foo, count + 1)
end
2) If bar(foo) is a function that returns a boolean:
defmodule Counter do
def count(foo) do
count(foo, 0)
end
defp count(foo, count) do
if bar(foo) do
count
else
count(foo, count + 1)
end
end
end
3) If bar(foo) returns something other than a boolean, that you can pattern-match on, like:
defmodule Counter do
def count(foo) do
count(foo, 0)
end
defp count(foo, count) do
case bar(foo) do
{:ok, baz} -> count
{:error, _} -> count(foo, count + 1)
end
end
end
Call the module and function:
Counter.count(foo)

Writing the function "once" in Elixir

I'm coming to Elixir from primarily a Javascript background. in JS, it's possible to write a higher order function "once" which returns a function that will invoke the passed in function only once, and returns the previous result on subsequent calls- the trick is manipulating variables that were captured via closure:
var once = (func) => {
var wasCalled = false, prevResult;
return (...args) => {
if (wasCalled) return prevResult;
wasCalled = true;
return prevResult = func(...args);
}
}
It seems to me that it's not possible to create this function in Elixir, due to its different variable rebinding behavior. Is there some other clever way to do it via pattern matching or recursion, or is it just not possible? Without macros that is, I'd imagine those might enable it. Thanks
Using the current process dictionary:
defmodule A do
def once(f) do
key = make_ref()
fn ->
case Process.get(key) do
{^key, val} -> val
nil ->
val = f.()
Process.put(key, {key, val})
val
end
end
end
end
Or if the function will be passed across processes, an ets table can be used:
# ... during application initialization
:ets.new(:cache, [:set, :public, :named_table])
defmodule A do
def once(f) do
key = make_ref()
fn ->
case :ets.lookup(:cache, key) do
[{^key, val}] -> val
[] ->
val = f.()
:ets.insert(:cache, {key, val})
val
end
end
end
end
Application.put_env / Application.get_env can also be used to hold global state, though usually is used for configuration settings.
It's not considered idiomatic in most cases, but you can do this with Agent:
defmodule A do
def once(fun) do
{:ok, agent} = Agent.start_link(fn -> nil end)
fn args ->
case Agent.get(agent, & &1) do
nil ->
result = apply(fun, args)
:ok = Agent.update(agent, fn _ -> {:ok, result} end)
result
{:ok, result} ->
result
end
end
end
end
Now if you run this:
once = A.once(fn sleep ->
:timer.sleep(sleep)
1 + 1
end)
IO.inspect once.([1000])
IO.inspect once.([1000])
IO.inspect once.([1000])
IO.inspect once.([1000])
You'll see that the first line is printed after 1 second, but the next 3 are printed instantly, because the result is fetched from the agent.
While both already given answers are perfectly valid, the most precise translation from your javascript is shown below:
defmodule M do
use GenServer
def start_link(_opts \\ []) do
GenServer.start_link(__MODULE__, nil, name: __MODULE__)
end
def init(_args) do
Process.sleep(1_000)
{:ok, 42}
end
def value() do
start_link()
GenServer.call(__MODULE__, :value)
end
def handle_call(:value, _from, state) do
{:reply, state, state}
end
end
(1..5) |> Enum.each(&IO.inspect(M.value(), label: to_string(&1)))
Use the same metric as in #Dogbert’s answer: the first value is printed with a delay, all subsequent are printed immediately.
This is an exact analog of your memoized function using GenServer stage. GenServer.start_link/3 returns one of the following:
{:ok, #PID<0.80.0>}
{:error, {:already_started, #PID<0.80.0>}}
That said, it is not restarted if it’s already started. I do not bother to check the returned value since we are all set in any case: if it’s the initial start, we call the heavy function, if we were already started, the vaklue is already at fingers in the state.

How to simulate Ruby's until loop in Elixir?

I'm converting a Ruby project to Elixir. How does Ruby's until loop translate to Elixir?
until scanner.eos? do
tokens << scan(line + 1)
end
Here's the full Ruby method:
def tokenize
#tokens = []
#lines.each_with_index do |text, line|
#scanner = StringScanner.new(text)
until #scanner.eos? do
#tokens << scan(line + 1)
end
end
#tokens
end
#lines is just a text file split by new lines. #lines = text.split("\n")
In Elixir, I've already converted the string scanner which looks like this: StringScanner.eos?(scanner):
#spec eos?(pid) :: boolean
def eos?(pid) when is_pid(pid) do
Also, in Elixir, tokens are tuples: #type token :: {:atom, any, {integer, integer}}. Where the {integer, integer} tuple is the line and position of the token.
This is the Elixir psuedo-code which doesn't quite work.
#spec scan(String.t, integer) :: token
def scan(text, line) when is_binary(text) and is_integer(line) do
string_scanner = StringScanner.new(text)
until StringScanner.eos?(string_scanner) do
result = Enum.find_value(#scanner_tokenizers, fn {scanner, tokenizer} ->
match = scanner.(string_scanner)
if match do
tokenizer.(string_scanner, match, line)
end
end)
IO.inspect result
end
StringScanner.stop(string_scanner)
result
end
Someone on the slack channel suggested using recursion, however they didn't elaborate with an example. I've seen recursion examples for summing / reducing which use accumulators etc. However, I don't see how that applies when evaluating a boolean.
Can anyone provide a working example which uses StringScanner.eos?(scanner)? Thanks.
It may be something like
def tokens(scanner) do
tokens(scanner, [])
end
defp tokens(scanner, acc) do
if StringScanner.eos?(scanner) do
acc
else
tokens(scanner, add_to_acc(scan_stuff(), acc))
end
end
At least this can be the general idea. As you'll see I kept a couple of functions very generic (scan_stuff/0 and add_to_acc/2) as I don't know how you mean to implement those; the first one is meant to do what scan(line + 1) does in the Ruby code, while the second one is meant to do what << does in the Ruby code (e.g., it could add the scanned stuff to the list of tokens or something similar).

How to convert a string to integer list in ocaml?

I need to pass two list as command line arguments in ocaml.
I used the following code to access it in the program.
let list1=Sys.argv.(1);;
let list2=Sys.argv.(2);;
I need to have the list1 and list2 as list of integers.
I am getting the error
This expression has type string but an expression was expected of type
int list
while processing.
How can I convert that arguments to a list of integers.
The arguments are passed in this format [1;2;3;4] [1;5;6;7]
Sys.argv.(n) will always be a string. You need to parse the string into a list of integers. You could try something like this:
$ ocaml
OCaml version 4.01.0
# #load "str.cma";;
# List.map int_of_string (Str.split (Str.regexp "[^0-9]+") "[1;5;6;7]");;
- : int list = [1; 5; 6; 7]
Of course this doesn't check the input for correct form. It just pulls out sequences of digits by brute force. To do better you need to do some real lexical analysis and simple parsing.
(Maybe this is obvious, but you could also test your function in the toplevel (the OCaml read-eval-print loop). The toplevel will handle the work of making a list from what you type in.)
As Sys.argv is a string array, you need to write your own transcription function.
I guess the simplest way to do this is to use the Genlex module provided by the standard library.
let lexer = Genlex.make_lexer ["["; ";"; "]"; ]
let list_of_string s =
let open Genlex in
let open Stream in
let stream = lexer (of_string s) in
let fail () = failwith "Malformed string" in
let rec aux acc =
match next stream with
| Int i ->
( match next stream with
| Kwd ";" -> aux (i::acc)
| Kwd "]" -> i::acc
| _ -> fail () )
| Kwd "]" -> acc
| _ -> fail ()
in
try
match next stream with
| Kwd "[" -> List.rev (aux [])
| _ -> fail ()
with Stream.Failure -> fail ()
let list1 = list_of_string Sys.argv.(1)
let list2 = list_of_string Sys.argv.(2)
Depending on the OCaml flavor you want to use, some other library may look more interesting. If you like yacc, Menhir may solve your problem in a few lines of code.

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