Simulating integer overflow in Python

Question

Python 2 has two integer datatypes int and long, and automatically converts between them as necessary, especially in order to avoid integer overflo

Answer 1

This function should convert your numbers to look like hardware integers. Depending on your application, you might need to apply this function between each stage of your operations.

def correct(value, bits, signed):
    base = 1 << bits
    value %= base
    return value - base if signed and value.bit_length() == bits else value

The following shortcut functions may come in handy for "casting" values to their appropriate range:

byte, sbyte, word, sword, dword, sdword, qword, sqword = (
    lambda v: correct(v, 8, False), lambda v: correct(v, 8, True),
    lambda v: correct(v, 16, False), lambda v: correct(v, 16, True),
    lambda v: correct(v, 32, False), lambda v: correct(v, 32, True),
    lambda v: correct(v, 64, False), lambda v: correct(v, 64, True)
)

As an example of how you might use them, a bug can reproduced that one might see in C. If one were to write a for loop using a byte to print out 0 - 255, the loop might never end. The following program demonstrates this problem:

#! /usr/bin/env python3
def main():
    counter = 0
    while counter < 256:
        print(counter)
        counter = byte(counter + 1)


def correct(value, bits, signed):
    base = 1 << bits
    value %= base
    return value - base if signed and value.bit_length() == bits else value


byte, sbyte, word, sword, dword, sdword, qword, sqword = (
    lambda v: correct(v, 8, False), lambda v: correct(v, 8, True),
    lambda v: correct(v, 16, False), lambda v: correct(v, 16, True),
    lambda v: correct(v, 32, False), lambda v: correct(v, 32, True),
    lambda v: correct(v, 64, False), lambda v: correct(v, 64, True)
)


if __name__ == '__main__':
    main()