Pythagorean calculation with sub-optimal numerics¶
ID: py/pythagorean Kind: problem Severity: warning Precision: medium Tags: - accuracy Query suites: - python-security-and-quality.qls
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
b**2 (or both) will. Thus, the calculation could overflow, even though the result is well within representable range.
sqrt(a**2 + b**2), use the built-in function
hypot(a,b) from the
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