Is the inverse of std::numeric_limits::infinity() zero?

Question

Is there anything in the C++ standard (or the IEEE 754 floating-point standard) that guarantees that 1./std::numeric_limits<double>::infinity() is zero (or at least a small number)?

[Very related](https://stackoverflow.com/questions/29426734/is-0-divided-by-infinity-guaranteed-to-be-0). — user202729, Aug 25 '18 at 14:46

John Zwinck · Answer 1 · 2018-08-26T02:03:27.353

15

Any finite number divided by infinity results in zero under IEEE 754 (and therefore the same in most typical C++ implementations).

If the sign of the of numerator and denominator differ, the result will be negative zero, which is equal to zero.

edited Aug 26 '18 at 02:03

answered Aug 25 '18 at 09:34

John Zwinck

239,568
38
324
436

3

Uhm, no, not exactly: if the finite number is negative it returns -0, which is not the same thing as 0 (although it compares equal). – Federico Poloni Aug 25 '18 at 12:31
6

Yeah, the proper wording would be "results in _a_ zero" :) – Ruslan Aug 25 '18 at 13:09
2

@FedericoPoloni: Both +0 and −0 in IEEE 754 specification level 2 (floating-point data) represent zero in level 1 (extended real numbers). – Eric Postpischil Aug 25 '18 at 17:46

score 11 · Accepted Answer · edited Aug 26 '18 at 11:05

11

Yes, according to the GNU C library reference manual (assuming IEEE 754):

Infinities propagate through calculations as one would expect: for example, 2 + ∞ = ∞, 4/∞ = 0

https://www.gnu.org/software/libc/manual/html_node/Infinity-and-NaN.html

You may want to check if your C++ compiler uses IEEE 754:

How to check if C++ compiler uses IEEE 754 floating point standard

edited Aug 26 '18 at 11:05

Peter Mortensen

30,738
21
105
131

answered Aug 25 '18 at 09:43

Gonen I

5,576
1
29
60

score 4 · Answer 3 · edited Jun 20 '20 at 09:12

IEEE 754-2008 6.1 says:

The behavior of infinity in floating-point arithmetic is derived from the limiting cases of real arithmetic with operands of arbitrarily large magnitude, when such a limit exists. Infinities shall be interpreted in the affine sense, that is: −∞ < {every finite number} < +∞.

Operations on infinite operands are usually exact and therefore signal no exceptions,…

Since the limit of 1/x as x increases without bound is zero, a consequence of this clause is that 1/∞ is zero.

Clause 6.3 tells us the sign of the result is +:

When neither the inputs nor result are NaN, the sign of a product or quotient is the exclusive OR of the operands’ signs;…

score 1 · Answer 4 · answered Aug 25 '18 at 16:50

1

if(std::numeric_limits<double>::is_iec559) yes(); else no();

(see 18.3.2.4)

IEC 559, which is identical with IEEE 754, guarantees that to be the case. However, C++ does not guarantee in any way that IEC 559 is in place (although 99.99% of the time that's just what happens to be the case, you still need to verify to be sure).

answered Aug 25 '18 at 16:50

Damon

67,688
20
135
185

1

Since nobody is likely to bother implementing and testing non-IEEE 754 platforms, one might as well `static_assert(std::numeric_limits::is_iec559)`. – John Zwinck Aug 26 '18 at 02:01
@JohnZwinck: That's actually a good point. If your code relies on IEEE 754 particulars, then it won't work if these aren't present. So `static_assert` is a very valid option (possibly, probably, even the better one). – Damon Aug 26 '18 at 12:26

Is the inverse of std::numeric_limits::infinity() zero?

4 Answers4

Linked