When solving a question of Project Euler I ran into the following logical error related to when n is updated.
while(n<1000):
#update n
#do something with n
#do stuff
vs
while(n<1000):
#do something with n
#do stuff
#update n
In the first instance, I ended up performing an operation with n even though the condition of n<1000 is violated.
Does this logical error have a name? How common is this bug?
I tried to look for it, I did find things about pre-incrementing and post-incrementing a variable. Although that is close to the error, it isn't exactly what is happening here. I found a reference to this in a SO answer about for loop vs while loop in the part describing how for loops are more concise and direct when compared to while loops. Essentially with while loops we end up running code after a variable update which could be buried somewhere in the code.
This is not always a bug: it depends on the algorithm. In some cases, you know that the original value of n is legal (so you can enter the loop), but you want to update and use the new value in your processing. You need to match your code to your algorithm. Your second code block is the canonical for-equivalent, and is more common.
This falls under the general heading of "off by 1 error".
Related
who can say me:
What is the difference between writing code inside else or just after the loops?
for x in range(6):
print('hello')
print('bye')
or
for x in range(6):
print('hello')
else:
print('bye')
From for/else documentation:
The else clause executes after the loop completes normally. This means that the loop did not encounter a break statement.
This means that in your specific example there is no difference in using else or not.
In the second example you're using conditional statement, Whereas in the first scenario after the loop completes, It automatically executes the next line.
Basically your post is an example of getting same result with alternative solutions.
Though your first solution can be considered as valid practice comparing with second one.
Although I would suggest you to read this article before posting any questions on stack overflow.
def function():
while True:
...omission...(this function is repeated permanently)
i =0
while i < 4:
driver.execute_script("""window.open("URL")""")
driver.switch_to.window(driver.window_handles[-1])
time.sleep(1)
function()
time.sleep(1)
i += 1 #open new tab and run function.
it doesn't work because while true loop is repeated permanently. Is there any ways to run multiple functions together?
https://imgur.com/a/4SIVekS This picture shows what I want
According to your picture, what you want is to launch the function a set number of times (4?), and run those in parrallel.
On a single core, as is the normal behavior, straight up parallel processing is impossible. You need to access other cores and manage a decentralized processing. while is useless there. I'm worried the level of difficulty is over your current skills, but here we go.
The overall flow that you (probably, depends on the actual memory safety of your functions) need is:
- to create a thread pool with the set number of threads for the number of runs you want.
- indicate the function you need to run
- start them, making sure the start itself is non-blocking.
- ensure one functions's processing doesn't impact another's results. race conditions are a common problem.
- gather results, again, in a non-blocking way.
You can use several methods. I highly recommend you read up a lot on the following documentations.
Threading:
https://docs.python.org/3/library/threading.html
Multiprocessing:
https://docs.python.org/3/library/multiprocessing.html
I don't understand your question because I don't understand what your function is supposed to do.
while True:
will always create an infinite loop. "while" is a command that tells python to loop through the following block so long as the expression following it evaluates to True. True always evaluates to True.
It seems like you want to use a conditional, like you do in "while x < 4".
x < 4
...is an expression that evaluates to true when x is less than 4, and false if x is not less than 4. Everything below the line:
while x < 4:
will then run if x is less than 4, and when it's done running that code, it will go back and evaluate if x is less than 4 again, and if it is, run the code again. To include another while loop inside of that loop, that new loop also needs an expression to evaluate. If you want to evaluate the same expression, write it out:
while x < 4:
# do something
while x < 4:
#do more things
# do even more things
# change x at some point, or else you're in an infinite loop.
However, there's no reason to do that specifically, because you're already doing it. All of the code is only running when x < 4, so checking that condition again right there is redundant, and doing it in another loop doesn't make sense. If the inside loop is also incrementing x, then the outside loop won't loop and doesn't need to increment x.
Also, if you want a function to check a condition based on a variable outside the function, you'll want to pass things to that function.
This code was written in Python 3. I am trying to find the 10001st prime number.
#10001st prime number
mylist=[]
def prime_index(n):
for i in range(99**99):
for x in range(2, int(i**1/2)):
if i % x == 0:
return False
return True
mylist.append(i)
n=int(n+1)
print(mylist[n])
break
prime_index(10001)
When I run, it says "list index out of range", referring to print(mylist[n]).
However I have been adding primes to the list along the way in the mylist.append(i). So can someone tell me what is the problem here because I do not understand what is going on. Is 99**99 too small? Or more subtle problem with code?
99**99 is not too small; if you actually print it, you're well beyond what you need (if you tried to run it out, you'd never finish, it's 657 bits of work). But your loop makes no sense; your inner loop will return either True or False immediately if it executes even once.
"Luckily" for you, it never executes even once. The first outer loop sets i to 0 the first time, so the inner loop doesn't run at all (side-note, you probably want i ** (1/2), not i ** 1 / 2; exponentiation has higher precedence than division). Because it doesn't run, you end up indexing into an empty list (and asking for index 10001 no less).
There are far too many problems in this code to address them all; look for other trial division prime finding code to get an idea of what it should look like.
The problem is that you try to print out the 10001st element as soon as you find (or not) the first prime. Also, note that you return from the routine without finding any prime number -- if you were to get that far. There is no way to reach the append statement.
You got to the print statement only because your first iteration has i = 0, so you don't enter the for loop at all.
Please follow the posting guidelines: take the time to research how to generate prime numbers. It can be much faster than you're doing, and will give you a nice, succinct bit of code to put into your program.
I have a piece of coding that is most basically a quiz. I had error handling so that if someone pressed a letter or anything other than a number, it would carry on with the quiz instead of letting the person attempt another question. (For example, if question 5 was 2+3 and they entered t, then it would carry on and not give them a different question for question 5).
I tried to update the coding, so it would loop:
def random_question():#defines function to get random question
count = 0#equates count to 0
for number in range (0,10):#makes the code generate the question 10 times
try:#the code makes it so the computer tries one thing and is the beggining
questionList = [random_sum,random_subtraction,random_times]#puts functions in a list to be called on
randomQuestion = random.choice(questionList)#calls on the list to generate the random questions
randomQuestion()#calls on variable
except ValueError:#if there is a value error or a type error it will notice it and try to stop it
print ("Please enter a number")#prints out the statement if there is a type error or a value error
else:
break
random_question()#calls on the function random_question
However, it comes up with a syntax error, and highlights the 'except' part next to the ValueError.
Any help as to why this is would be much appreciated.
Your except statement should have the same indention as your try statement. In your sample code it is indented an extra tab which would cause that error.
Python takes indention seriously and it is often the culprit. I'm unclear if your def line is a place holder or if this code is part of it, but if this code is part of the function definition you have other indention issues to worry about.
I'd suggest going through it carefully and making sure everything is lined up properly. The basic rule of thumb is, if something is part of something else, it is indented under it.
def something():
step 1
step 2
if (condition):
the true thing
else:
the false thing
while (something):
repeat something
this is not part of the function anymore
I am learning Python 3 and working on an exercise that calls for writing a Python program which simulates/reads a BASIC program as input. I am stuck on writing the part of the Python program that should detect infinite loops. Here is the code I have so far:
def execute(prog):
while True:
location = 0
if prog[location] == len(prog) - 1:
break
return "success"
getT = prog[location].split()
T = len(getT) - 1
location = findLine(prog, T)
visited = [False] * len(prog)
Here, prog is a list of strings containing the BASIC program (strings are in the form of 5 GOTO 30, 10 GOTO 20, etc.).
T is the target string indicated in prog[location].
If the BASIC program has an infinite loop, then my Python program will have an infinite loop. I know that if any line is visited twice, then it loops forever, and my program should return "infinite loop".
A hint given by the tutorial assistant says "initialize a list visited = [False] * len(prog) and change visited[i] to True when prog[i] is visited. Each time through the loop, one value updates in visited[]. Think about how you change a single value in a list. Then think about how you identify which value in visited[] needs to change."
So this is the part I am stuck on. How do I keep track of which strings in prog have been visited/looped through?
I'm not sure I agree that visiting a line twice proves an infinite loop. See the comments under the question. But I can answer the actual question.
Here's the hint:
A hint given by the tutorial assistant says "initialize a list visited = [False] * len(prog) and change visited[i] to True when prog[i] is visited. Each time through the loop, one value updates in visited[]. Think about how you change a single value in a list. Then think about how you identify which value in visited[] needs to change."
This is saying you should have two lists, one that contains the program, and one that contains true/false flags. The second one is to be named visited and initially contains False values.
The Python code is just like the hint says:
visited = [False] * len(prog)
This uses the * list operator, "list repetition", to repeat a length-1 list and make a new list of a longer length.
To change visited[i] to True is simple:
visited[i] = True
Then you can do something like this:
if visited[i]:
print("We have already visited line {}".format(i))
print("Infinite loop? Exiting.")
sys.exit(1)
Note that we are testing for the True value by simply saying if visited[i]:
We could also write if visited[i] == True: but the shorter form is sufficient and is customary in the Python community. This and other customary idioms are documented here: http://python.net/~goodger/projects/pycon/2007/idiomatic/handout.html
For a program this small, it's not too bad to keep two lists like this. For larger and complex programs, I prefer to keep everything together in one place. This would use a "class" which you might not have learned yet. Something like this:
class ProgramCode(object):
def __init__(self, statement):
self.code = statement
self.visited = False
prog = []
with open(input_basic_program_file, "rt") as f:
for line in f:
prog.append(ProgramCode(line))
Now instead of two lists, we have a single list where each item is a bit of BASIC code and a visited flag.
P.S. The above shows an explicit for loop that repeatedly uses .append() to add to a list. An experienced Python developer would likely use a "list comprehension" instead, but I wanted to make this as easy to follow as possible.
Here's the list comprehension. Don't worry if it looks weird now; your class will teach this to you eventually.
with open(input_basic_program_file, "rt") as f:
prog = [ProgramCode(line) for line in f]
I know of no automatic way of infinite loop detection in Python, but by using divide and conquer methods and testing individual functions, you can find the offending function or block of code and then proceed to debug further.
If the Python program outputs data, but you never see that output, that's a good indicator you have an infinite loop. You can test all your functions in the repl, and the function that does "not come back" [to the command prompt] is a likely suspect.
You can write output under a debug variable of some sort, to be shut off when everything works. This could be a member variable of a Python class to which your code would have to have access to at any time, or you could have a module-scoped variable like Debug=1 and use debug levels to print varying amounts of debug info, like 1 a little, 2 more, 3, even more, and 4 verbose.
As an example, if you printed the value of a loop counter in a suspected function, then eventually that loop counter would keep printing well beyond the count of data (test records) you were using to test.
Here is a combination I came up with using parts of J. Carlos P.'s answer with the hints that steveha gave and using the hint that the instructions gave:
def execute(prog):
location = 0
visited = [False] * len(prog)
while True:
if location==len(prog)-1:
return "success"
findT = prog[location].split()
T = findT[- 1]
if visited[location]:
return "infinite loop"
visited[location] = True
location = findLine(prog, T)