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submitted 1 year ago* (last edited 1 year ago) by andioop@programming.dev to c/programming_horror@programming.dev
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[-] lysdexic@programming.dev 1 points 1 year ago

The decimal representation of real numbers isn’t unique, so this could tell me that “2 = 1.9999…” is odd.

I don't think your belief holds water. By definition an even number, once divided by 2, maps to an integer. In binary representations, this is equivalent to a right shift. You do not get a rounding error or decimal parts.

But this is nitpicking a tongue-in-cheek comment.

[-] Chobbes@lemmy.world 1 points 1 year ago

“1.99999…” is an integer, though! If you’re computing with arbitrary real numbers and serializing it to a string, how do you know to print “2” instead of “1.9999…”? This shouldn’t be decidable, naively if you have a program that prints “1.” and then repeatedly runs a step of an arbitrary Turing machine and then prints “9” if it did not terminate and stops printing otherwise, determining if the number being printed would be equal to 2 would solve the halting problem.

Arbitrary precision real numbers are not represented by finite binary integers. Also a right shift on a normal binary integer cannot tell you if the number is even. A right shift is only division by 2 on even numbers, otherwise it’s division by 2 rounded down to the nearest integer. But if you have a binary integer and you want to know if it’s even you can just check the least significant bit.

this post was submitted on 07 Dec 2023
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