I am learning Julia, and in particular I am trying to get a grasp of macros, so I found, amongst other things Some useful macros for Julia - Github, and in the process of deciphering them I got this behavior in the REPL :
julia> macro once_then(expr::Expr)
#assert expr.head == :while
esc(quote
$(expr.args[2]) # body of loop
$expr # loop
end)
end
#once_then (macro with 1 method)
julia> i = 0
0
julia> #once_then while i < 10
i += 1
end
ERROR: UndefVarError: i not defined
Stacktrace:
[1] macro expansion at ./REPL[34]:2 [inlined]
[2] top-level scope at ./REPL[31]:5
julia> i
1
It clearly had access to i in the first iteration of the loop, since it incremented it, but then did i become Undefed somewhere between the end of the first loop and the beginning of the second ?
From what I can see the parenthesis after esc should include everything up to the end of the loop...
I literally just copy-pasted the code into the terminal, and I get the same behavior from the #until macro found at Julia language - Until loop, so I don't think the problem would come from the code itself... is there something crucial I'm missing ?
(btw, I'm running 1.0.4, so it shouldn't be an issue of backwards compatibility...)
Your macro is fine. the problem is the scope of the variables involved. (for more information, look at this: JuliaLang - Scope of Variables
in a nutshell, the variable i in i = 0 is in a global scope, where the loop is in a local scope. in the REPL, you can add the keywork global this to make your code work:
julia> #once_then while i < 10
global i += 1
end
other option is to do all this in a function, so all variables have local scope:
function fn()
i = 0
#once_then while i < 10
i += 1
end
return i
end
Related
I don't understand why the recursive function always gives me zero result, even if I put values inside the array.
it seems that size (a) == 0
recursive function binarySearch_R (a, value) result (bsresult)
real, intent(in) :: a(6), value
integer :: bsresult, mid
mid = size(a)/2 + 1
if (size(a) == 0) then
bsresult = 0 ! not found
else if (a(mid) > value) then
bsresult= binarySearch_R(a(:mid-1), value)
else if (a(mid) < value) then
bsresult = binarySearch_R(a(mid+1:), value)
if (bsresult /= 0) then
bsresult = mid + bsresult
end if
else
bsresult = mid ! SUCCESS!!
end if
end function binarySearch_R
program hji
read*, a
read*, value
print*, binarySearch_R
end program hji
Chapter 1: The dangers of implicit typing
The first thing I strongly recommend you do is to include the line
implicit none
after the program line. This will suppress implicit typing, and the resulting errors will give you some useful insight into what is happening.
If you did that, you'd get an error message:
$ gfortran -o binsearch binsearch.f90
binsearch.f90:23:12:
read*, a
1
Error: Symbol ‘a’ at (1) has no IMPLICIT type
binsearch.f90:27:25:
print*,binarySearch_R
1
Error: Symbol ‘binarysearch_r’ at (1) has no IMPLICIT type
binsearch.f90:24:16:
read*, value
1
Error: Symbol ‘value’ at (1) has no IMPLICIT type
It doesn't matter that a, value, and binarySearch_R were defined in the function. As the function is not part of the program block, the program doesn't know what these are.
With implicit typing active, it simply assumed that all three are simple real variables. (The type depends on the first letter of the variable name, i through n are integer, everything else is real)
Because this implicit typing can so easily hide coding errors, it's strongly, strongly suggested to always switch it off.
Which also means that we have to declare the variables a and value in the program:
program hji
implicit none
real :: a(6), value
...
end program hji
Chapter 2: How to introduce a function to the program?
So how does the program get access to the function? There are four ways:
The best way: Use a module
module mod_binsearch
implicit none
contains
recursive function binarySearch_R (a, value) result (bsresult)
...
end function binarySearch_R
end module mod_binsearch
program hji
use mod_binsearch
implicit none
real :: a(6), value
...
end program hji
Note that the use statement has to be before the implicit none.
This method leaves the function separate, but callable.
It automatically checks that the parameters (that's something we'll be coming to in a bit) are correct.
Have the function contained in the program.
Between the final line of code of the program and the end program statement, add the keyword contains, followed by the function code (everything from recursive function ... to end function ...).
This is the quick-and-dirty method. You have to be careful with this method as the function will automatically have access to the program's variables unless there's a new variable with that name declared inside the function.
The convoluted way: Interfaces
Create an interface block in the declaration section of your program's source code,
and repeat the interface information in there.
This still allows the compiler to check whether the function is invoked correctly, but it's up to you to ensure that this interface block is correct and matches the actual implementation.
The really, really ugly way: Declare it like a variable, invoke it like a function.
Please don't do that.
Chapter 3: Calling a function
When you call a function, you have to use the parentheses and give it all the parameters that it expects. In your case, you need to type
print *, binarySearch_r(a, value)
Chapter 4: Dynamic arrays as dummy parameters
In the successive recursive calls to the function, the array gets smaller and smaller.
But the dummy parameter is always the same size (6). Not only will this interfere with your algorithm, but this can also lead to dangerously undefined memory access.
Fortunately, specially for intent(in) dummy parameters, you can use dynamic arrays:
recursive function binarySearch_R(a, value)
real, intent(in) :: a(:), value
The single colon tells the compiler to expect a one-dimensional array, but not the length of it. Since you're already using size(a), it should automatically work.
Too long for a comment, but not an answer (and to any Fortran experts reading this, yes, there are one or two places where I gloss over some details because I think they are unimportant at this stage) ...
The way the code is written does not allow the compiler to help you. As far as the compiler is concerned there is no connection between the function and the program. As far as the program is concerned a is, because you haven't told the compiler otherwise, assumed to be a real scalar value. The a in the program is not the same thing as the a in the function - there is no connection between the function and the program.
The same is true for value.
The same is true for binarysearch_r - and if you don't believe this delete the function definition from the source code and recompile the program.
So, what must you do to fix the code ?
First step: modify your source code so that it looks like this:
program hji
... program code goes here ...
contains
recursive function binarySearch_R (a, value) result (bsresult)
... function code goes here ...
end function binarySearch_R
end program hji
This first step allows the compiler to see the connection between the program and the function.
Second step: insert the line implicit none immediately after the line program hji. This second step allows the compiler to spot any errors you make with the types (real or integer, etc) and ranks (scalar, array, etc) of the variables you declare.
Third step: recompile and start dealing with the errors the compiler identifies. One of them will be that you do not pass the arguments to the function so the line
print*, binarySearch_R
in the program will have to change to
print*, binarySearch_R(a, value)
This is a weird one print return 9 and then it prints 1 , also i checked debugger in pycharm and the (stuff) keeps counting down for some reason
def repeater(stuff):
if stuff != 9:
stuff += 1
print(stuff)
repeater(stuff)
return stuff
print(repeater(0))
When you call repeater(stuff), you're not passing the variable stuff, you're passing a copy of the variable stuff. When you say stuff += 1, you're not modifying the stuff you called the function with, you're modifying a copy of it. That change isn't reflected in the original when you exit the function.
Then, when the function exits, you don't do anything with the returned value of stuff - which is, again, copied out of the function in reverse. Python does let you call the function without using its return value, but it looks like your intent here is to apply the returned value of repeater(stuff) to stuff.
To accomplish that, simply change the line
repeater(stuff)
to
stuff = repeater(stuff)
The reason for that additional 1 that is coming in the end is that repeater(stuff) is returning the value of stuff which is being received by your print statement i.e. print(repeater(0)). When all the recursive calls return back, none of the values are stored/used but the very first call that was made by print(repeater(0)) obtains a value of 1 because repeater(stuff) returns the value of stuff which would be 1 after stuff += 1 during the first call.
You can read more about how recursion works for more clarity.
I am trying to solve this problem of displaying prime numbers upto a specified number.
While I have the standard solution for it. I first tried to solve it myself and ended up writing this code. But I'm not getting the desired output. The standard solution uses the while loop as the main loop. And if that's the ideal loop for this example, why would that be so ?
def count_primes(num):
my_primes=[2]
if num<2:
return 0
for x in range(3,num+1,2):
for y in range(3,x+1):
if x%y==0:
break
else:
my_primes.append(x)
return my_primes
count_primes(100)
I expected a list of all prime numbers up to 100. Instead the output displayed only [2]. My guess is that the 'break' keyword broke out of the entire loop instead of only the if loop.
In your second for loop change range(3,x+1) to range(3,x), since your loop goes until y=x and x%x is always 0.
This is a follow-up to my previous question.
I am finally able to reproduce the error here:
my #recentList = prompt("Get recentList: e.g. 1 2 3: ").words || (2,4,6);
say "the list is: ", #recentList;
for #recentList -> $x {
say "one element is: ", $x;
say "element type is: ", $x.WHAT;
say "test (1,2,3).tail(\"2\") : ", (1,2,3).tail("2");
say ( (10.rand.Int xx 10) xx 15 ).map: { #($_.tail($x)); };
}
And the results are ok as long as I use the default list by just hitting return at the prompt and not entering anything. But if I enter a number, it gives this error:
Get recentList: e.g. 1 2 3: 2
the list is: [2]
one element is: 2
element type is: (Str)
test (1,2,3).tail("2") : (2 3)
This type cannot unbox to a native integer: P6opaque, Str
in block at intType.p6 line 9
in block <unit> at intType.p6 line 5
If tail("2") works, why does tail($x) fail? Also, in my original code, tail($x.Int) wouldn't correct the problem, but it did here.
This is at best a nanswer. It is a thus-far failed attempt to figure out this problem. I may have just wandered off into the weeds. But I'll publish what I have. If nothing else, maybe it can serve as a reminder that the first three steps below are sensible ones; thereafter I'm gambling on my ability to work my way forward by spelunking source code when I would probably make much faster and more reliable progress by directly debugging the compiler as discussed in the third step.
OK, the first step was an MRE. What you've provided was an E that was fully R and sufficiently M. :)
Step #2 was increasing the M (golfing). I got it down to:
Any.tail('0'); # OK
Any.tail('1'); # BOOM
Note that it can be actual values:
1.tail('1'); # BOOM
(1..2).tail('1'); # BOOM
But some values work:
(1,2).tail('1'); # OK
Step #3 probably should be to follow the instructions in Playing with the code of Rakudo Perl 6 to track the compiler's execution, eg by sticking says in its source code and recompiling it.
You may also want to try out App::MoarVM::Debug. (I haven't.)
Using these approaches you'll have the power to track with absolute precision what the compiler does for any code you throw at it. I recommend you do this even though I didn't. Maybe you can figure out where I've gone wrong.
In the following I trace this problem by just directly spelunking the Rakudo compiler's source code.
A search for "method tail" in the Rakudo sources yielded 4 matches. For my golf the matching method is a match in core/AnyIterableMethods.pm6.
The tail parameter $n clearly isn't a Callable so the pertinent line that continues our spelunking is Rakudo::Iterator.LastNValues(self.iterator,$n,'tail').
A search for this leads to this method in core/Iterator.pm6.
This in turn calls this .new routine.
These three lines:
nqp::if(
n <= 0, # must be HLL comparison
Rakudo::Iterator.Empty, # negative is just nothing
explain why '0' works. The <= operator coerces its operands to numeric before doing the numeric comparison. So '0' coerces to 0, the condition is True, the result is Rakudo::Iterator.Empty, and the Any.tail('0') yields () and doesn't complain.
The code that immediately follows the above three lines is the else branch of the nqp::if. It closes with nqp::create(self)!SET-SELF(iterator,n,f).
That in turn calls the !SET-SELF routine, which has the line:
($!lastn := nqp::setelems(nqp::list, $!size = size)),
Which attempts to assign size, which in our BOOM case is '1', to $!size. But $!size is declared as:
has int $!size;
Bingo.
Or is it? I don't know if I really have correctly tracked the problem down. I'm only spelunking the code in the github repo, not actually running an instrumented version of the compiler and tracing its execution, as discussed as the sensible step #3 for trying to figure out the problem you've encountered.
Worse, when I'm running a compiler it's an old one whereas the code I'm spelunking is the master...
Why does this work?
(*,*).tail('1') # OK
The code path for this will presumably be this method. The parameter $n isn't a Callable so the code path will run thru the path that uses the $n in the lines:
nqp::unless(
nqp::istype($n,Whatever) || $n == Inf,
$iterator.skip-at-least(nqp::elems($!reified) - $n.Int)
The $n == Inf shouldn't be a problem. The == will coerce its operands to numerics and that should take care of $n being '1'.
The nqp::elems($!reified) - $n.Int shouldn't be a problem either.
The nqp ops doc shows that nqp::elems always returns an int. So this boils down to an int - Int which should work.
Hmm.
A blame of these lines shows that the .Int in the last line was only added 3 months ago.
So, clutching at straws, what happens if one tries:
(my int $foo = 1) - '1' # OK
Nope, that's not the problem.
It seems the trail has grown cold or rather I've wandered off the actual execution path.
I'll publish what I've got. Maybe someone else can pick it up from here or I'll have another go in a day or three...
I am struggling with re-assigning a variable in a loop in Julia. I have a following example:
infile = "test.txt"
feature = "----"
for ln in 1:3
println(feature)
feature = "+"
end
open(infile) do f
#if true
# println(feature)
# feature = "----"
println(feature)
for ln in 1:5 #eachline(f)
println("feature")
#fails here
println(feature)
# because of this line:
feature = "+"
end
end
It fails if I do reassignment within the loop. I see that the issue with the variable scope, and because nested scopes are involved. The reference says that the loops introduce 'soft' scope. I cannot find out from the manual what scope open expression belongs to, but seemingly it screws things up, as if I replace open with if true, things run smoothly.
Do I understand correctly that open introduces 'hard' scope and that it is the reason why re-assignment retroactively makes the variable undefined?
You should think of
open("file") do f
...
end
as
open(function (f)
...
end, "file")
that is, do introduces the same kind of hard scope as a function or -> would.
So to be able to write to feature from the function, you need to do
open(infile) do f
global feature # this means: use the global `feature`
println(feature)
for ln in 1:5
println("feature")
println(feature)
feature = "+"
end
end
Note that this is only the case in top-level (module) scope; once inside a function, there are no hard scopes.
(The for loop in this case is a red herring; regardless of the soft scope of the loop, the access to feature will be limited by the hard scope of the anonymous function introduced by do.)