Programming Forums - View Single Post - KeyboardInterrupt when calling a function

Vahagn · Feb 1st, 2006, 3:01 PM

Hi, I am toying with the following program to generate prime numbers:

import sys
from math import sqrt, floor

def prime_gen(count):

    num = 2
    ct = 0  
    while ct < count:
        for i in range(2, int(floor(sqrt(num)))):
            if num % i == 0:             
                num = num + 1
            elif num % i != 0:
                print num
                ct = ct + 1
                num = num + 1               
         
prime_gen(100) 
#x = int(sys.argv[1])
#prime_gen(x)

When I call prime_gen at the prompt (=execute the program), the prompt freezes and I have to exit it with Ctrl+C. The error message says there is a KeyboardInterrupt and the lines 17 and 9 cause the problem, I however do not understand WHAT the problem is - thanks beforehand for help.

V.

Feb 1st, 2006, 3:01 PM	#1
Vahagn Newbie Join Date: Feb 2006 Posts: 5 Rep Power: 0	KeyboardInterrupt when calling a function Hi, I am toying with the following program to generate prime numbers: (Toggle Plain Text) import sys from math import sqrt, floor def prime_gen(count): num = 2 ct = 0 while ct < count: for i in range(2, int(floor(sqrt(num)))): if num % i == 0: num = num + 1 elif num % i != 0: print num ct = ct + 1 num = num + 1 prime_gen(100) #x = int(sys.argv[1]) #prime_gen(x) import sys from math import sqrt, floor def prime_gen(count): num = 2 ct = 0 while ct < count: for i in range(2, int(floor(sqrt(num)))): if num % i == 0: num = num + 1 elif num % i != 0: print num ct = ct + 1 num = num + 1 prime_gen(100) #x = int(sys.argv[1]) #prime_gen(x) When I call prime_gen at the prompt (=execute the program), the prompt freezes and I have to exit it with Ctrl+C. The error message says there is a KeyboardInterrupt and the lines 17 and 9 cause the problem, I however do not understand WHAT the problem is - thanks beforehand for help. V.