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1 | 1 | using OrdinaryDiffEq, ParameterizedFunctions, |
2 | | - DiffEqBase, RecursiveArrayTools |
| 2 | + DiffEqBase, RecursiveArrayTools, Unitful, StaticArrays |
3 | 3 | using Base.Test |
4 | 4 |
|
5 | 5 | # Here's the problem to solve |
@@ -35,3 +35,18 @@ ans = [[1 2; 3 4],[1 4; 4 4.5],[5 7; 4.5 4.5]] |
35 | 35 | ans = [[2.333333333333 4.666666666666; 3.6666666666666 6.0], [2.3333333 3.0; 5.0 2.6666666]] |
36 | 36 | @test recursive_mean(B,2)[1] ≈ ans[1] |
37 | 37 | @test recursive_mean(B,2)[2] ≈ ans[2] |
| 38 | + |
| 39 | +A = zeros(5,5) |
| 40 | +recursive_unitless_eltype(A) == Float64 |
| 41 | +A = zeros(5,5)*1u"kg" |
| 42 | +recursive_unitless_eltype(A) == Float64 |
| 43 | +AA = [zeros(5,5) for i in 1:5] |
| 44 | +recursive_unitless_eltype(AA) == Array{Float64,2} |
| 45 | +AofA = [copy(A) for i in 1:5] |
| 46 | +recursive_unitless_eltype(AofA) == Array{Float64,2} |
| 47 | +AofSA = [@SVector [2.0,3.0] for i in 1:5] |
| 48 | +recursive_unitless_eltype(AofSA) == SVector{2,Float64} |
| 49 | +AofuSA = [@SVector [2.0u"kg",3.0u"kg"] for i in 1:5] |
| 50 | +recursive_unitless_eltype(AofuSA) == SVector{2,Float64} |
| 51 | + |
| 52 | +@inferred recursive_unitless_eltype(AofuSA) |
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