Exit while loop in Python

Question

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

Answer 1

The while loop will match the condition only when the control returns back to it, ie 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).

Answer 2

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

Answer 3

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.

Answer 4

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

Answer 5

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

Answer 6

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