Haskell: TCO and Lazy evaluation - haskell

I'm trying to understand why the first main does not terminate when c is not valid, while the second terminates. From the description here
main is just an unevaluated thunk, and executing it is just building up the data structure. I'm trying to apply the same principle here and see why the first main does not terminate. If someone can help me understand this part, or give me pointers to understanding this would be great. Apart from this, why is GHCI not able to recognize this as TCO ? does is not fit the definition?
main = loop
where
loop = do
c <- getChar
case valid c of
Nothing -> return ()
Just b -> print b
print c
loop
> main :: IO ()
> main = loop
> where
> loop = do
> c <- getChar
> case validate c of
> Nothing -> return ()
> Just b -> do
> print b
> loop
Thanks.

Tail-call optimization has nothing to do with this behaviour. The problem is simply that the first code contains an infinite loop while the second doesn't.
Your first code is similar to the imperative (python-like):
def loop():
c = getChar()
if valid c:
do_something()
else:
do_something_else()
print(c)
loop()
While the latter is similar to:
def loop():
c = getChar()
if valid c:
do_something()
else:
do_something_else()
print(c)
loop()
Note how in the latter case the call to loop() is inside the branch else while in the former it's outside and thus is called at every loop invocation.
Also note that return in Haskell does not terminate the function calls. It's just an IO action that has a certain value and no side-effects.
For example:
main = do
c <- return 1
print c
In the above code the return does not prevent the output from print.

Related

Why along with output it is showing NONE each time? [duplicate]

What does the return statement do? How should it be used in Python?
How does return differ from print?
See also
Often, people try to use print in a loop inside a function in order to see multiple values, and want to be able to use the results from outside. They need to be returned, but return exits the function the first time. See How can I use `return` to get back multiple values from a loop? Can I put them in a list?.
Often, beginners will write a function that ultimately prints something rather than returning it, and then also try to print the result, resulting in an unexpected None. See Why is "None" printed after my function's output?.
Occasionally in 3.x, people try to assign the result of print to a name, or use it in another expression, like input(print('prompt:')). In 3.x, print is a function, so this is not a syntax error, but it returns None rather than what was displayed. See Why does the print function return None?.
Occasionally, people write code that tries to print the result from a recursive call, rather than returning it properly. Just as if the function were merely called, this does not work to propagate the value back through the recursion. See Why does my recursive function return None?.
Consider How do I get a result (output) from a function? How can I use the result later? for questions that are simply about how to use return, without considering print.
The print() function writes, i.e., "prints", a string in the console. The return statement causes your function to exit and hand back a value to its caller. The point of functions in general is to take in inputs and return something. The return statement is used when a function is ready to return a value to its caller.
For example, here's a function utilizing both print() and return:
def foo():
print("hello from inside of foo")
return 1
Now you can run code that calls foo, like so:
if __name__ == '__main__':
print("going to call foo")
x = foo()
print("called foo")
print("foo returned " + str(x))
If you run this as a script (e.g. a .py file) as opposed to in the Python interpreter, you will get the following output:
going to call foo
hello from inside foo
called foo
foo returned 1
I hope this makes it clearer. The interpreter writes return values to the console so I can see why somebody could be confused.
Here's another example from the interpreter that demonstrates that:
>>> def foo():
... print("hello within foo")
... return 1
...
>>> foo()
hello within foo
1
>>> def bar():
... return 10 * foo()
...
>>> bar()
hello within foo
10
You can see that when foo() is called from bar(), 1 isn't written to the console. Instead it is used to calculate the value returned from bar().
print() is a function that causes a side effect (it writes a string in the console), but execution resumes with the next statement. return causes the function to stop executing and hand a value back to whatever called it.
Think of the print statement as causing a side-effect, it makes your function write some text out to the user, but it can't be used by another function.
I'll attempt to explain this better with some examples, and a couple definitions from Wikipedia.
Here is the definition of a function from Wikipedia
A function, in mathematics, associates one quantity, the argument of the function, also known as the input, with another quantity, the value of the function, also known as the output..
Think about that for a second. What does it mean when you say the function has a value?
What it means is that you can actually substitute the value of a function with a normal value! (Assuming the two values are the same type of value)
Why would you want that you ask?
What about other functions that may accept the same type of value as an input?
def square(n):
return n * n
def add_one(n):
return n + 1
print square(12)
# square(12) is the same as writing 144
print add_one(square(12))
print add_one(144)
#These both have the same output
There is a fancy mathematical term for functions that only depend on their inputs to produce their outputs: Referential Transparency. Again, a definition from Wikipedia.
Referential transparency and referential opaqueness are properties of parts of computer programs. An expression is said to be referentially transparent if it can be replaced with its value without changing the behavior of a program
It might be a bit hard to grasp what this means if you're just new to programming, but I think you will get it after some experimentation.
In general though, you can do things like print in a function, and you can also have a return statement at the end.
Just remember that when you use return you are basically saying "A call to this function is the same as writing the value that gets returned"
Python will actually insert a return value for you if you decline to put in your own, it's called "None", and it's a special type that simply means nothing, or null.
I think the dictionary is your best reference here
Return and Print
In short:
return gives something back or replies to the caller of the function while print produces text
In python, we start defining a function with def, and generally - but not necessarily - end the function with return.
Suppose we want a function that adds 2 to the input value x. In mathematics, we might write something like f(x) = x + 2, describing that relationship: the value of the function, evaluated at x, is equal to x + 2.
In Python, it looks like this instead:
def f(x):
return x + 2
That is: we define a function named f, which will be given an x value. When the code runs we figure out x + 2, and return that value. Instead of describing a relationship, we lay out steps that must be taken to calculate the result.
After defining the function, it can be called with whatever argument you like. It doesn't have to be named x in the calling code, and it doesn't even have to be a variable:
print f(2)
>>> 4
We could write the code for the function in some other ways. For example:
def f(x):
y = x + 2
return y
or even
def f(x):
x = x + 2
return x
Again, we are following steps in order - x = x + 2 changes what x refers to (now it means the result from the sum), and that is what gets returned by return x (because that's the value *at the time that the return happens).
return means "output this value from this function".
print means "send this value to (generally) stdout"
In the Python REPL, a function's return value will be output to the screen by default (this isn't the same as printing it). This output only happens at the REPL, not when running code from a .py file. It is the same as the output from any other expression at the REPL.
This is an example of print:
>>> n = "foo\nbar" #just assigning a variable. No output
>>> n #the value is output, but it is in a "raw form"
'foo\nbar'
>>> print(n) #the \n is now a newline
foo
bar
>>>
This is an example of return:
>>> def getN():
... return "foo\nbar"
...
>>> getN() #When this isn't assigned to something, it is just output
'foo\nbar'
>>> n = getN() # assigning a variable to the return value. No output
>>> n #the value is output, but it is in a "raw form"
'foo\nbar'
>>> print(n) #the \n is now a newline
foo
bar
>>>
This answer goes over some of the cases that have not been discussed above.
The return statement allows you to terminate the execution of a function before you reach the end. This causes the flow of execution to immediately return to the caller.
In line number 4:
def ret(n):
if n > 9:
temp = "two digits"
return temp #Line 4
else:
temp = "one digit"
return temp #Line 8
print("return statement")
ret(10)
After the conditional statement gets executed the ret() function gets terminated due to return temp (line 4).
Thus the print("return statement") does not get executed.
Output:
two digits
This code that appears after the conditional statements, or the place the flow of control cannot reach, is the dead code.
Returning Values
In lines number 4 and 8, the return statement is being used to return the value of a temporary variable after the condition has been executed.
To bring out the difference between print and return:
def ret(n):
if n > 9:
print("two digits")
return "two digits"
else :
print("one digit")
return "one digit"
ret(25)
Output:
two digits
'two digits'
Note that return can also be used for control flow. By putting one or more return statements in the middle of a function, we can say: "stop executing this function. We've either got what we wanted or something's gone wrong!"
For example, imagine trying to implement str.find(sub) if we only had str.index(sub) available (index raises a ValueError if the substring isn't found, whereas find returns -1).
We could use a try/except block:
def find(s: str, sub: str) -> int:
try:
return s.index(sub)
except ValueError:
return -1
This is fine, and it works, but it's not very expressive. It's not immediately clear what would cause str.index to raise a ValueError: a reader of this code must understand the workings of str.index in order to understand the logic of find.
Rather than add a doc-string, saying "...unless sub isn't found, in which case return -1", we could make the code document itself, like this:
def find(s: str, sub: str) -> int:
if sub not in s:
return -1
return s.index(sub)
This makes the logic very clear.
The other nice thing about this is that once we get to return s.index(sub) we don't need to wrap it in a try/except because we already know that the substring is present!
See the Code Style section of the Python Guide for more advice on this way of using return.
To put it as simply as possible:
return makes the value (a variable, often) available for use by the caller (for example, to be stored by a function that the function using return is within). Without return, your value or variable wouldn't be available for the caller to store/re-use.
print, by contrast, prints to the screen - but does not make the value or variable available for use by the caller.
Difference between "return" and "print" can also be found in the following example:
RETURN:
def bigger(a, b):
if a > b:
return a
elif a <b:
return b
else:
return a
The above code will give correct results for all inputs.
PRINT:
def bigger(a, b):
if a > b:
print a
elif a <b:
print b
else:
print a
NOTE: This will fail for many test cases.
ERROR:
----
FAILURE: Test case input: 3, 8.
Expected result: 8
FAILURE: Test case input: 4, 3.
Expected result: 4
FAILURE: Test case input: 3, 3.
Expected result: 3
You passed 0 out of 3 test cases
Here is my understanding. (hope it will help someone and it's correct).
def count_number_of(x):
count = 0
for item in x:
if item == "what_you_look_for":
count = count + 1
return count
So this simple piece of code counts number of occurrences of something. The placement of return is significant. It tells your program where do you need the value. So when you print, you send output to the screen. When you return you tell the value to go somewhere. In this case you can see that count = 0 is indented with return - we want the value (count + 1) to replace 0.
If you try to follow logic of the code when you indent the return command further the output will always be 1, because we would never tell the initial count to change.
I hope I got it right.
Oh, and return is always inside a function.
return should be used for recursive functions/methods or you want to use the returned value for later applications in your algorithm.
print should be used when you want to display a meaningful and desired output to the user and you don't want to clutter the screen with intermediate results that the user is not interested in, although they are helpful for debugging your code.
The following code shows how to use return and print properly:
def fact(x):
if x < 2:
return 1
return x * fact(x - 1)
print(fact(5))
This explanation is true for all of the programming languages not just python.
return is part of a function definition, while print outputs text to the standard output (usually the console).
A function is a procedure accepting parameters and returning a value. return is for the latter, while the former is done with def.
Example:
def timestwo(x):
return x*2
Best thing about return function is you can return a value from function but you can do same with print so whats the difference ?
Basically return not about just returning it gives output in object form so that we can save that return value from function to any variable but we can't do with print because its same like stdout/cout in C Programming.
Follow below code for better understanding
CODE
def add(a, b):
print "ADDING %d + %d" % (a, b)
return a + b
def subtract(a, b):
print "SUBTRACTING %d - %d" % (a, b)
return a - b
def multiply(a, b):
print "MULTIPLYING %d * %d" % (a, b)
return a * b
def divide(a, b):
print "DIVIDING %d / %d" % (a, b)
return a / b
print "Let's do some math with just functions!"
age = add(30, 5)
height = subtract(78, 4)
weight = multiply(90, 2)
iq = divide(100, 2)
print "Age: %d, Height: %d, Weight: %d, IQ: %d" % (age, height, weight, iq)
# A puzzle for the extra credit, type it in anyway.
print "Here is a puzzle."
what = add(age, subtract(height, multiply(weight, divide(iq, 2))))
print "That becomes: ", what, "Can you do it by hand?"
We are now doing our own math functions for add, subtract, multiply, and divide. The important thing to notice is the last line where we say return a + b (in add). What this does is the following:
Our function is called with two arguments: a and b.
We print out what our function is doing, in this case "ADDING."
Then we tell Python to do something kind of backward: we return the addition of a + b. You might say this as, "I add a and b then return them."
Python adds the two numbers. Then when the function ends, any line that runs it will be able to assign this a + b result to a variable.
The simple truth is that print and return have nothing to do with each other. print is used to display things in the terminal (for command-line programs).1 return is used to get a result back when you call a function, so that you can use it in the next step of the program's logic.
Many beginners are confused when they try out code at Python's interpreter prompt2, like
>>> def example():
... return 1
...
>>> example()
1
The value was displayed; doesn't this mean that return displays things? No. If you try the same code in a .py file, you can see for yourself that running the script doesn't cause the 1 to display.
This shouldn't actually be confusing, because it works the same way as any other expression:
>>> 1 + 1
2
This displays at the interactive prompt, but not if we make a script that just says 1 + 1 and try running it.
Again: if you need something to display as part of your script, print it. If you need to use it in the next step of the calculation, return it.
The secret is that the interactive prompt is causing the result to be displayed, not the code. It's a separate step that the prompt does for you, so that you can see how the code works a step at a time, for testing purposes.
Now, let's see what happens with print:
>>> def example():
... return 'test'
...
>>> print(example())
test
The result will display, whether we have this in an interactive prompt or in a script. print is explicitly used to display the value - and as we can see, it displays differently. The interactive prompt uses what is called the repr of the value that was returned from example, while print uses the str of the value.
In practical terms: print shows us what the value looks like, in text form (for a string, that just means the contents of the string as-is). The interactive prompt shows us what the value is - typically, by writing something that looks like the source code we would use to create it.3
But wait - print is a function, right? (In 3.x, anyway). So it returned a value, right? Isn't the interpreter prompt supposed to display that in its separate step? What happened?
There is one more trick: print returns the special value None, which the interpreter prompt will ignore. We can test this by using some expressions that evaluate to None:
>>> None
>>> [None][0]
>>> def example():
... pass # see footnote 4
...
>>> example()
>>>
In each case, there is no separate line at all for output, not even a blank line - the interpreter prompt just goes back to the prompt.
1 It can also be used to write into files, although this is a less common idea and normally it will be clearer to use the .write method.
2 This is sometimes called the REPL, which stands for "read-eval-print loop".
3 This isn't always practical, or even possible - especially once we start defining our own classes. The firm rule is that repr will lean on the .__repr__ method of the object to do the dirty work; similarly, str leans on .__str__.
4 Functions in Python implicitly return None if they don't explicitly return a value.
Return statement -- will return some values according your function.
def example(n):
if n == 5:
return true
else:
return false
if you call above function and you pass number 5 then it will return true else it will return false.
Printing function -- it will print content that you have given to the print function or with in print function bracket.
def example(n):
if n == 5:
print("number is equal")
else:
print("number is not equal")

Undefined variable error not reported using multi-threading

The following code is a simplification of a code of a library I'm working on. The code runs but it does not produces the expected output. The problem turns out to be the use of a wrong variable name (k instead of c) in the function fn. However, no error was not reported.
fn(ch, c) = put!(ch, k)
chnl = Channel{Char}(1)
function producer(ch,str)
for c in str
fn(ch, c)
end
end
consumer(ch) = for c in ch
print(c)
end
Threads.#spawn producer(chnl, "Hello World!\n")
Threads.#spawn consumer(chnl)
sleep(1)
println("Done")
Detecting this problem took me a while on the original code, because no error was reported. Is there something I could use to detect this type of problem in the future, specially because function fn is suppose to be written by the user?
seems to produce error for me:
julia> Threads.#spawn producer(chnl, "Hello World!\n")
Task (failed) #0x00007fccc68cc160
UndefVarError: k not defined
changing this to:
t1 = Threads.#spawn producer(chnl, "Hello World!\n")
t2 = Threads.#spawn consumer(chnl)
wait(t1)
wait(t2)
println("Done")
running as a script:
akako#ThinkX /tmp> julia test.jl
ERROR: LoadError: TaskFailedException
Stacktrace:
[1] wait(t::Task)
# Base ./task.jl:322
[2] top-level scope
# /tmp/test.jl:17
nested task error: UndefVarError: k not defined
Stacktrace:
[1] fn(ch::Channel{Char}, c::Char)
# Main /tmp/test.jl:1
[2] producer(ch::Channel{Char}, str::String)
# Main /tmp/test.jl:7
[3] (::var"#1#2")()
# Main ./threadingconstructs.jl:169
After digging a little bit more, an alternative solution is to wrap the body of the function producer in a try-catch statement
function producer(ch,str)
try
for c in str
fn(ch, c)
end
catch
for (exc, bt) in Base.catch_stack()
showerror(stdout, exc, bt)
println(stdout)
end
exit()
end
end
Resulting in a stack race of the problem:
/tmp> julia problem.jl
UndefVarError: k not defined
Stacktrace:
[1] fn(ch::Channel{Char}, c::Char)
# Main /tmp/problem.jl:1
[2] producer(ch::Channel{Char}, str::String)
# Main /tmp/problem.jl:8
[3] (::var"#1#2")()
# Main ./threadingconstructs.jl:169

Search of element from List in the String

I have a list like List = ["google","facebook","instagram"] and a string P1 = "https://www.google.co.in/webhp?pws=0&gl=us&gws_rd=cr".
Now I need to find which element of List is present inside the P1.
For this I implemented below recursive function, but it returns ok as final value, is there a way that when (in this case) google is found, then H is returned and terminate the other recursive calls in stack.
I want this function to return google.
traverse_list([],P1)-> ok;
traverse_list([H|T],P1) ->
Pos=string:str(P1,H),
if Pos > 1 ->
io:fwrite("Bool inside no match is ~p~n",[Pos]),
io:fwrite("inside bool nomathc, ~p~n",[H]),
H;
true->
io:fwrite("value found :: ~p~n",[Pos])
end,
traverse_list(T,P1).
It returns ok because the stop condition of your recursion loop does it:
traverse_list([],P1)-> ok;
For this you should use lists:filter/2 or a list comprehension:
List = ["google","facebook","instagram"],
P1 = "https://www.google.co.in/webhp?pws=0&gl=us&gws_rd=cr",
lists:filter(fun(X) -> string:str(P1,X) > 1 end,List),
% or
[X || X <- List, string:str(P1,X) > 1],

Combining Scalaz IO monad with Stream for simple echo program using takeWhile method

I'm looking for a solution for "Life, the Universe, and Everything" problem on spoj.com using Scalaz IO Monad and Stream.
The problem is to rewrite small numbers from input to output and to stop processing input after reading in the number 42.
I created following classes:
import scalaz._, effect._
def ReadInt: IO[Int] = IO {
readInt
}
def PrintInt(i: Int): IO[Unit] = IO {
println(i)
}
def EchoInt: IO[Int] = {
for {
i <- ReadInt
_ <- PrintInt(i)
} yield (i)
}
Never-ending loop works as expected:
scala> import Scalaz._
scala> Stream.continually(EchoInt).sequence.unsafePerformIO
12
22
32
42
52
However, when I want it to finish on the number 42 it doesn't finish as well:
scala> Stream.continually(EchoInt).sequence.map(_.takeWhile(_ != 42)).unsafePerformIO
12
22
32
42
52
I know that this code is supposed to print 42 after all (as opposed to problem statement), but I wanted to simplify the example code.
Where am I making mistake?
Stream.continually(EchoInt).sequence repeats EchoInt infinite times, so your program is basically an infinite loop.
What you want to do is to put some sort of break into your loop, which will analyze the input on each iteration and stop the loop if the input equals 42. There is no break in functional programming (because there are no loops, only recursion), so you simply do not continue the recursion:
import scalaz.syntax.monad._
...
def echoInts: IO[Unit] = {
for {
i <- EchoInt
_ <- echoInts.whenM(i != 42) // Break if i == 42
} yield ()
}
echoInts.unsafePerformIO()

Scala Comprehension Errors

I am working on some of the exercism.io exercises. The current one I am working on is for Scala DNA exercise. Here is my code and the errors that I am receiving:
For reference, DNA is instantiated with a strand String. This DNA can call count (which counts the strand for the single nucleotide passed) and nucletideCounts which counts all of the respective occurrences of each nucleotide in the strand and returns a Map[Char,Int].
class DNA(strand:String) {
def count(nucleotide:Char): Int = {
strand.count(_ == nucleotide)
}
def nucleotideCounts = (
for {
n <- strand
c <- count(n)
} yield (n, c)
).toMap
}
The errors I am receiving are:
Error:(10, 17) value map is not a member of Int
c <- count(n)
^
Error:(12, 5) Cannot prove that Char <:< (T, U). ).toMap
^
Error:(12, 5) not enough arguments for method toMap: (implicit ev:
<:<[Char,(T, U)])scala.collection.immutable.Map[T,U]. Unspecified
value parameter ev. ).toMap
^
I am quite new to Scala, so any enlightenment on why these errors are occurring and suggestions to fixing them would be greatly appreciated.
for comprehensions work over Traversable's that have flatMap and map methods defined, as the error message is pointing out.
In your case count returns with a simple integer so no need to "iterate" over it, just simply add it to your result set.
for {
n <- strand
} yield (n, count(n))
On a side note this solution is not too optimal as in the case of a strand AAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAAA count is going to be called many times. I would recommend calling toSet so you get the distinct Chars only:
for {
n <- strand.toSet
} yield (n, count(n))
In line with Akos's approach, consider a parallel traversal of a given strand (String),
strand.distinct.par.map( n => n -> count(n) )
Here we use distinct to gather unique items and construct each Map association in map.
A pipeline solution would look like:
def nucleotideCounts() = strand.groupBy(identity).mapValues(_.length)
Another approach is
Map() ++ {for (n <- strand; c = count(n)) yield n->c}
Not sure why it's different than {...}.toMap() but it gets the job done!
Another way to go is
Map() ++ {for (n <- strand; c <- Seq(count(n))) yield n->c}

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