In Python, when a int bigger than 2**31, then it will turn to a long:
a = 2147483647
a + 1 = 2147483648
b = -2147483648
b - 1 = -2147483649
but I need the Python int overflow like the int in C:
a = 2147483647
a + 1 = -2147483648
b = -2147483648
b - 1 = 2147483647
Is it possible? thanks in advance!
Try numpy:
>>> x = numpy.int32(2147483647)
>>> x
2147483647
>>> type(x)
<type 'numpy.int32'>
>>> x+1
__main__:1: RuntimeWarning: overflow encountered in long_scalars
-2147483648
>>> type(x+1)
<type 'numpy.int32'>
Just make sure to call int on these things before passing them to code that expects normal Python overflow behavior.
You can define your own class and override the __int__() special method, along with various other mathematical operator special methods, to emulate a numeric type. Then your class can maintain the invariant that the value is always in the proper range.
For example:
def class Int32:
def __init__(self):
self.value = 0
def __init__(self, value):
# Wrap value into [-2**31, 2**31-1]
self.value = (value + 2**31) % 2**32 - 2**31
def __int__(self):
return self.value
def __add__(self, other):
return Int32(self.value + other.value)
# ... etc. for other mathematical operators
