max(.NaN, x) != max(x, .NaN)

Question

Created Jun ’15

Replies 5

Boosts 0

Participants 2

import Darwin

print(  max(.NaN, 1.23) ) // 1.23
print(  max(1.23, .NaN) ) // nan

print( fmax(.NaN, 1.23) ) // 1.23
print( fmax(1.23, .NaN) ) // 1.23

print(  max(.NaN, 1.0) !=  max(1.0, .NaN) ) // true
print( fmax(.NaN, 1.0) != fmax(1.0, .NaN) ) // false

Why should the order of the arguments decide if nan is propagated or not? And why should max not behave as fmax?

I haven't checked what IEEE-754 says, but the man page for fmax is pretty clear about this:

"If exactly one argument is a NaN, fmax() returns the other argument. If both arguments are NaNs, fmax() returns a NaN."

The reason for why Swift's standard library max (and min) are behaving this way is probably because they are defined something like this:

func myMax<T : Comparable>(x: T, _ y: T) -> T { return x > y ? x : y }
func myMin<T : Comparable>(x: T, _ y: T) -> T { return x > y ? y : x }

Which produces the same results (ie order of arguments decides whether nan is propagated).

Btw, apparently Haskell (GHCi) has this "problem" too, but the order that results in propagation is the opposite(!), so maybe I should just keep my mouth shut:

Prelude> let nan = 0/0
Prelude> nan
NaN
Prelude> max nan 1.23
NaN
Prelude> max 1.23 nan
1.23

But still, the simple bold question above begs for an answer.

Boost

Answer 1

tim1724 OP

Jun ’15

File a radar. I'm sure this isn't intentional behavior, but a side effect of min()/max() being defined as generics, as you suggested. They clearly need to provide an alternate implemenation for floating point types.

0

Answer 2

Jens OP

Jun ’15

I filed one (18371906) nine months ago, it is one of the few that is still open, hence this thread : ).

0

Answer 3

tim1724 OP

Jun ’15

The workaround for now is to declare your own:

func max<T : FloatingPointType>(x: T, _ y: T) -> T {
    if x.isNaN { return y }
    if y.isNaN { return x }
    return x > y ? x : y
}
func min<T : FloatingPointType>(x: T, _ y: T) -> T {
    if x.isNaN { return y }
    if y.isNaN {return x }
    return x < y ? x : y
}

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Answer 4

tim1724 OP

Jun ’15

Wow. That's bizarre. Especially given how trivial it would be for them to fix.

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Answer 5

tim1724 OP

Jun ’15

And now I've filed my own bug report: 21432613

0