To be clear, I am not looking for NaN or infinity, or asking what the answer to x/0 should be. What I’m looking for is this:
Based on how division is performed in hardware (I do not know how it is done), if division were to be performed with a divisor of 0, and the processor just chugged along happily through the operation, what would come out of it?
I realize this is highly dependent on the dividend, so for a concrete answer I ask this: What would a computer spit out if it followed its standard division operation on 42 / 0?
Update:
I’ll try to be a little clearer. I’m asking about the actual operations done with the numbers at the bit level to reach a solution. The result of the operation is just bits. NaN and errors/exceptions come into play when the divisor is discovered to be zero. If the division actually happened, what bits would come out?
It might just not halt. Integer division can be carried out in linear time through repeated subtraction: for 7/2, you can subtract 2 from 7 a total of 3 times, so that’s the quotient, and the remainder (modulus) is 1. If you were to supply a dividend of 0 to an algorithm like that, unless there were a mechanism in place to prevent it, the algorithm would not halt: you can subtract 0 from 42 an infinite number of times without ever getting anywhere.
From a type perspective, this should be intuitive. The result of an undefined computation or a non-halting one is ⊥ (“bottom”), the undefined value inhabiting every type. Division by zero is not defined on the integers, so it should rightfully produce ⊥ by raising an error or failing to terminate. The former is probably preferable. 😉
Other, more efficient (logarithmic time) division algorithms rely on series that converge to the quotient; for a dividend of 0, as far as I can tell, these will either fail to converge (i.e., fail to terminate) or produce 0. See Division on Wikipedia.
Floating-point division similarly needs a special case: to divide two floats, subtract their exponents and integer-divide their significands. Same underlying algorithm, same problem. That’s why there are representations in IEEE-754 for positive and negative infinity, as well as signed zero and NaN (for 0/0).