How is dividing by 0 not a 'divide by zero' exception?

Question

This code produces a System.ArgumentOutOfRangeException because numberOfIntervals is infinity. But why doesn't this produce a divide by zero exception? The calculation ends up being -524 / 0. Shouldn't that produce a divide by zero exception?

DateTime startDate = new DateTime(2019, 1, 1);
DateTime referenceDate = new DateTime(2020, 6, 8);

double numberOfIntervals = (startDate - referenceDate).TotalDays / 0;

Console.WriteLine(numberOfIntervals);

Does this answer your question? (you divide a double by an int) https://stackoverflow.com/questions/44258124/divide-by-zero-and-no-error/44258269 — DasElias, Jun 11 '20 at 17:00
The numerator is a double. So it becomes the max positive number for a double. — jdweng, Jun 11 '20 at 17:02
It partially answers my question. That's helpful, but *why* not throw `divide by zero` for doubles as well? — Bob Horn, Jun 11 '20 at 17:03
Sometimes in floating-point arithmetic it's useful to be able to represent INFINITY... So `1f / 0 == double.PositiveInfinity` will be True — Stewart Ritchie, Jun 11 '20 at 17:06

score 3 · Accepted Answer · answered Jun 11 '20 at 17:09

3

The implementation is done by IEEE 754 standard.

Division by zero: an operation on finite operands gives an exact infinite result, e.g., 1/0 or log(0). By default, returns ±infinity.

answered Jun 11 '20 at 17:09

Sir Rufo

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I see that in the exception handling section: https://en.wikipedia.org/wiki/IEEE_754#Exception_handling I guess I'm still wondering why treat double differently than int. – Bob Horn Jun 11 '20 at 17:14
1

Maybe this can help https://en.wikipedia.org/wiki/Division_by_zero – Sir Rufo Jun 11 '20 at 17:18

How is dividing by 0 not a 'divide by zero' exception?

1 Answers1