Convert NaN to correct type in sqrt (#64)

* Convert NaN to correct type in sqrt

* Split sqrt into AbstractFloat and Real case

Author: DanielVandH

src/NaNMath.jl

@@ -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)

test/runtests.jl

@@ -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]))