strange behavior of math.sqrt() in new 3.0 version

David David
Fri Dec 26 16:52:24 EST 2008


I'm a newbee trying 3.0   Please help with  math.sqrt() 

At the command line this function works correctly 
      >>> import math
              n = input("enter a number > ")
              s = math.sqrt(n)
     An entry of 9 or 9.0  will yield 3.0

Yet the same code in a script gives an error message
     Script1
                   import math
                   n = input("enter a number > ")
                   s = math.sqrt(n)
              
               Traceback (most recent call last) :
                  File "<stdin>", line 1, in <module>
                  File "script1.py" line 3 in <module>
                     s = math.sqrt(n)
               TypeError : a float is required
     Entering 9 or 9.0 gives same error message.

   According to the math module the results of all
   functions are floats.  However it says nothing about
   inputs.

Strangely the above code runs fine in version 2.5  ( ? )
and will handle large integers.

I've read the documentation for 3.0 including the section
"Floating Point Arithmetic: Issues & Limitations" and it
helps nada.



More information about the Python-list mailing list