Pythagorean calculation with sub-optimal numerics¶
ID: py/pythagorean
Kind: problem
Security severity:
Severity: warning
Precision: medium
Tags:
- accuracy
Query suites:
- python-security-and-quality.qls
Click to see the query in the CodeQL repository
Calculating the length of the hypotenuse using the standard formula c = sqrt(a**2 + b**2)
may lead to overflow if the two other sides are both very large. Even though c
will not be much bigger than max(a, b)
, either a**2
or b**2
(or both) will. Thus, the calculation could overflow, even though the result is well within representable range.
Recommendation¶
Rather than sqrt(a**2 + b**2)
, use the built-in function hypot(a,b)
from the math
library.
Example¶
The following code shows two different ways of computing the hypotenuse. The first is a direct rewrite of the Pythagorean theorem, the second uses the built-in function.
# We know that a^2 + b^2 = c^2, and wish to use this to compute c
from math import sqrt, hypot
a = 3e154 # a^2 > 1e308
b = 4e154 # b^2 > 1e308
# with these, c = 5e154 which is less that 1e308
def longSideDirect():
return sqrt(a**2 + b**2) # this will overflow
def longSideBuiltin():
return hypot(a, b) # better to use built-in function
References¶
Python Language Reference: The hypot function
Wikipedia: Hypot.