[wtallis]

"a+b+c" doesn't describe a unique evaluation order. You need some parentheses to disambiguate which changes are due to associativity vs commutativity. a+(b+c)=(c+b)+a should be true of floating point numbers, due to commutativity. a+(b+c)=(a+b)+c may fail due to the lack of associativity.

[adastra22]

It is not, due to precision. Consider a=1.00000, b=-0.99999, and c=0.00000582618.

[jcranmer]

No, the two evaluations will give you exactly the same result: https://play.rust-lang.org/?version=stable&mode=debug&editio...

IEEE 754 operations are nonassociative, but they are commutative (at least if you ignore the effect of NaN payloads).

[imtringued]

https://play.rust-lang.org/?version=stable&mode=debug&editio...

You're supposed to do (a+b) to demonstrate the effect, because floating point subtraction that results in a number near zero is sensitive to rounding (worst case, a non-zero number gets you a zero number), which can introduce a huge error when a and b are very similar numbers.

[jcranmer]

When you go from (a + b) + c to a + (b + c), you're invoking the associative property, not the commutative property.

The confusion between associativity and commutativity is the entire point of this thread!

[dbdr]

Is there a case involving NaN where they are not commutative? Do you mean getting a different bit-level representation of NaN?

[jcranmer]

IEEE 754 doesn't (usually) distinguish between different NaN encodings for the purposes of semantics--if the result is a NaN, it doesn't specify which NaN the result is. Most hardware vendors implement a form of NaN propagation: when both inputs are NaN, one of the operands is returned, for example, always the left NaN is returned if both are NaN.

As a side note: all compilers I'm aware of make almost no guarantees on preserving the value of NaN payloads, hence they consider floating-point operations to be fully commutative, and there's no general way to guarantee that they evaluate in exactly the order you specified.

[Remnant44]

In practical use for simd, various min/max operations. On Intel at least, they propagate nan or not based on operand order

[boulos]

Also all comparisons.

[immibis]

Does (1.00000+-0.99999)+0.00000582618 != 0.00000582618+(-0.99999+1.00000) ? This would disprove commutativity. But I think they're equal.

[zygentoma]

You still need to specify an evaluation order …