9

In the code below, I'd like the while loop to exit as soon as a + b + c = 1000. However, testing with print statements shows that it just continues until the for loops are done. I've tried while True and then in the if statement set False but that results in an infinite loop. I thought using x = 0 and then setting x = 1 might work but that too just runs until the for loops finish. What is the most graceful and fastest way to exit? Thanks.

a = 3
b = 4
c = 5
x = 0
while x != 1:
    for a in range(3,500):
        for b in range(a+1,500):
            c = (a**2 + b**2)**0.5
            if a + b + c == 1000:
                print a, b, c
                print a*b*c
                x = 1
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2
  • 3
    ignoring the loop issue... you're going to have floating point issues here... specifically, c is a floating point # and so you're going to want to either: round it, cast it, or check if abs(a+b+c-1000) < 0.00001 (or some other epsilon). Or I could be totally mistaken. Commented May 20, 2013 at 18:58
  • c is floating, but it didn't seem to matter. The program finds an a+b+c=1000 so apparently doesn't care about 1000 vs 1000.0. Good to know though because it could be an issue at times. Commented May 21, 2013 at 21:28

6 Answers 6

Reset to default
10

The while loop will match the condition only when the control returns back to it, i.e when the for loops are executed completely. So, that's why your program doesn't exits immediately even though the condition was met.

But, in case the condition was not met for any values of a,b,c then your code will end up in an infinite loop.

You should use a function here as the return statement will do what you're asking for.

def func(a,b,c):
    for a in range(3,500):
        for b in range(a+1,500):
            c = (a**2 + b**2)**0.5
            if a + b + c == 1000:
                print a, b, c
                print a*b*c
                return # causes your function to exit, and return a value to caller

func(3,4,5)

Apart from @Sukrit Kalra's answer, where he used exit flags you can also use sys.exit() if your program doesn't have any code after that code block.

import sys
a = 3
b = 4
c = 5
for a in range(3,500):
    for b in range(a+1,500):
        c = (a**2 + b**2)**0.5
        if a + b + c == 1000:
            print a, b, c
            print a*b*c
            sys.exit()     #stops the script

help on sys.exit:

>>> print sys.exit.__doc__
exit([status])

Exit the interpreter by raising SystemExit(status).
If the status is omitted or None, it defaults to zero (i.e., success).
If the status is numeric, it will be used as the system exit status.
If it is another kind of object, it will be printed and the system
exit status will be one (i.e., failure).
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8 Comments

Thank you. If I use a function I don't really need the while loop at all. As I understand, the function would run until the condition to return is met and then it would exit. That works fine, but doesn't really help me understand why my original while loop won't exit.
It exits. It just does some extra for loops.
your code will stop executing when the for loops are finished. In the OP's post, however, if the if condition was not met in any case, the loops will start over from the beginning.
That's the issue, I'd like it to exit without doing the extra for loops. Basically, I'm trying to figure out how to make a while loop exit early once a condition is met. Doing it outside a function doesn't seem to work. Maybe I should be using functions more, but seems like it should work.
@caadrider then why did you use a while loop?
|
6

If you don't want to make a function ( which you should and refer to Ashwini's answer in that case), here is an alternate implementation.

>>> x = True
>>> for a in range(3,500):
        for b in range(a+1, 500):
            c = (a**2 + b**2)**0.5
            if a + b + c == 1000:
                 print a, b, c
                 print a*b*c
                 x = False
                 break
         if x == False:
            break
200 375 425.0
31875000.0
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6 Comments

That works. I wish there was a way to break out of nested loops in one fell swoop. I thought the while loop would provide that option but so far I can't figure out how to make it work.
You won't be able to swiftly break out of all the nested loops. If you're in a script however, you can use sys.exit() which will break out of the complete execution of the file and execute nothing after it.
@caadrider Python doesn't have a goto statement like C, so you can't just jump out of nested loops suddenly. So a good option here is to use functions.
@Ashwini - thanks for the clarification. Looks like functions are the best way to do this. Time to embrace them. :)
This PEP might be a useful read.
|
1

You can refactor the inner code into a function and use return to exit:

def inner():
    for a in range(3,500):
        for b in range(a+1,500):
            c = (a**2 + b**2)**0.5
            if a + b + c == 1000:
                print a, b, c
                print a*b*c
                return False
    return True

while inner():
    pass

Have a look at this question.

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Comments

1

The problem is, even though you set x=1 when a+b+c==1000, you do not break out of the two for loops when that condition is met, and so the while loop doesn't know that x==1 until both for loops finish. To avoid this problem, you can add explicit break statements to the for loops (and as Sukrit Kalra points out, the while loop becomes unnecessary).

a = 3
b = 4
c = 5
x = 0
for a in range(3,500):
  for b in range(a+1,500):
     c = (a**2 + b**2)**0.5
     if a + b + c == 1000:
        print a, b, c
        print a*b*c
        x = 1
        break
  if x==1:
     break
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1 Comment

You won't actually need the while loop in this case.
1

You could use a break statement:

a = 3
b = 4
c = 5
x = 0
while x != 1:
    for a in range(3,500):
        for b in range(a+1,500):
            c = (a**2 + b**2)**0.5
            if a + b + c == 1000:
                print a, b, c
                print a*b*c
                break
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Comments

0

You can wrap with try/excep and raise when the condition is met.

class FinitoException(Exception):
    pass

a = 3
b = 4
c = 5
x = 0
try:
  for a in range(3,500):
      for b in range(a+1,500):
          c = (a**2 + b**2)**0.5
          if a + b + c == 1000:
              print a, b, c
              print a*b*c
              raise FinitoException()
except FinitoException:
    return # or whatever
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1 Comment

I'll have to wait until my Python gets a lot better for this one. Thanks for the tip though. I'll look into try statements. Haven't learned those yet.

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