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Convert NaN to correct type in sqrt #64

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Merged
merged 2 commits into from
Jan 14, 2023
Merged

Convert NaN to correct type in sqrt #64

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42ba274
Convert NaN to correct type in sqrt
DanielVandH Jan 10, 2023
2727977
Split sqrt into AbstractFloat and Real case
DanielVandH Jan 14, 2023
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3 changes: 2 additions & 1 deletion src/NaNMath.jl
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Expand Up @@ -14,7 +14,8 @@ end

# Would be more efficient to remove the domain check in Base.sqrt(),
# but this doesn't seem easy to do.
sqrt(x::Real) = x < 0.0 ? NaN : Base.sqrt(x)
sqrt(x::T) where {T<:AbstractFloat} = x < 0.0 ? T(NaN) : Base.sqrt(x)
sqrt(x::Real) = sqrt(float(x))

# Don't override built-in ^ operator
pow(x::Float64, y::Float64) = ccall((:pow,libm), Float64, (Float64,Float64), x, y)
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9 changes: 9 additions & 0 deletions test/runtests.jl
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Expand Up @@ -18,6 +18,15 @@ using Test
@test NaNMath.pow(-1.5,2.3) isa Float64
@test isnan(NaNMath.sqrt(-5))
@test NaNMath.sqrt(5) == Base.sqrt(5)
@test isnan(NaNMath.sqrt(-3.2f0)) && NaNMath.sqrt(-3.2f0) isa Float32
@test isnan(NaNMath.sqrt(-BigFloat(7.0))) && NaNMath.sqrt(-BigFloat(7.0)) isa BigFloat
@test isnan(NaNMath.sqrt(-7)) && NaNMath.sqrt(-7) isa Float64
@inferred NaNMath.sqrt(5)
@inferred NaNMath.sqrt(5.0)
@inferred NaNMath.sqrt(5.0f0)
@inferred NaNMath.sqrt(-5)
@inferred NaNMath.sqrt(-5.0)
@inferred NaNMath.sqrt(-5.0f0)
@test NaNMath.sum([1., 2., NaN]) == 3.0
@test NaNMath.sum([1. 2.; NaN 1.]) == 4.0
@test isnan(NaNMath.sum([NaN, NaN]))
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