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0 pointssva_1mo ago0 comments

> floating point accumulation doesn't commute

It is commutative (except for NaN). It isn't associative though.

0 comments

ekelsen1mo ago

I think it commutes even when one or both inputs are NaN? The output is always NaN.

addaon1mo ago

NaNs are distinguishable. /Which/ NaN you get doesn't commute.

ekelsen1mo ago

I guess at the bit level, but not at the level of computation? Anything that relies on bit patterns of nans behaving in a certain way (like how they propagate) is in dangerous territory.

1 more reply

DavidVoid1mo ago

Unless you compile with fast-math ofc, because then the compiler will assume that NaN never occurs in the program.

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0 pointssva_1mo ago0 comments

> floating point accumulation doesn't commute

It is commutative (except for NaN). It isn't associative though.

0 comments

ekelsen1mo ago

I think it commutes even when one or both inputs are NaN? The output is always NaN.

addaon1mo ago

NaNs are distinguishable. /Which/ NaN you get doesn't commute.

ekelsen1mo ago

I guess at the bit level, but not at the level of computation? Anything that relies on bit patterns of nans behaving in a certain way (like how they propagate) is in dangerous territory.

1 more reply

DavidVoid1mo ago

Unless you compile with fast-math ofc, because then the compiler will assume that NaN never occurs in the program.

j / k navigate · click thread line to collapse