numpy - casting uint32 to uint8

Question

I have the following code in python using numpy:

import numpy
a = 423
b = numpy.uint8(a)
print(b)

It gives me the result:

b = 167

I understand that uint8 (unsigned 8-bit integer) can represent values from 0 to 255. What I don't understand is how numpy does the cut-off or "truncation" of the value 423 to the value 167. Subtracting 255 from 423 (423 - 255) gives me 168 which is off by one. Anybody knows what is the formula used by numpy in numpy.uint8 when the input value is 255 or greater?

Answer 1

Don't think of it as subtraction, more an application of mod. Simply put, numpy.uint8 (currently) truncates all the bits after the first 8 (i.e. the number 256) starting from the Least Significant Bit and returns the resulting number. Take a look below:

>>> bin(423) 
'0b110100111'
>>> bin(numpy.uint8(423))
'0b10100111'
>>> >>> bin(12345)
'0b11000000111001'
>>> bin(numpy.uint8(12345))
'0b111001'