I am wondering how to define inf and -inf as an int in Python 2.7. I tried and it seems inf and -inf only work as a float.
a = float('-inf') # works b = float('inf') # works c = int('-inf') # compile error, ValueError: invalid literal for int() with base 10: 'inf' d = int('inf') # compile error, ValueError: invalid literal for int() with base 10: 'inf' To summarise what was said in the comments
There is no way to represent infinity as an integer in Python. This matches the behaviour of many other languages. However, due to Python's dynamic typing system, you can use float('inf') in place of an integer, and in most situations it will behave as you would expect.
As far as creating a 'double' for infinity, in Python there is just one floating point type, called float, unlike other languages such as Java which uses the term float and double for floating point numbers with different precision. In Python, floating point numbers usually use double-precision, so they act the same as doubles in Java.
math.inf (having put import math in your imports), and negative infinity as -math.inf.
math.inf is now better than writing float('inf') as suggested in this answer. However, it gives the same actual value, it doesn't answer the original question any better. I hadn't meant to suggest that it does, but I can see that my comment didn't make that clear.Something very big that is a integer (made in 3.x, not tested in 2.x):
0x40000 #will most likely crash the shell so if you wanted a infinite variable you would do:
Infinity = 0x40000 You can do any Integer to Hexedecimal here:
https://www.binaryhexconverter.com/decimal-to-hex-converter
(make sure to add the 0x before the value it returns for python)
0x40001 is larger than 0x40000, so it is not a good idea to use 0x40000 as "infinity".
float('inf')