I’m wondering how to use @inbounds when vectorizing a function with dot notation or @.. I found Base.@propagate_inbounds but am not sure I’m using it correctly. Consider
using BenchmarkTools
Base.@propagate_inbounds active_dot(x, l, u, δ) = (x ≤ l + δ) || (x ≥ u - δ)
function active!(A, x, l, u, δ)
@inbounds for i in eachindex(x)
A[i] = (x[i] ≤ l[i] + δ[i]) || (x[i] ≥ u[i] - δ[i])
end
A
end
function foo(n)
l = zeros(n)
u = ones(n)
x = rand(n) - 0.5
rtol = atol = 1.0e-8
# is it possible to write the following with dot notation??
δ = [-Inf < l[i] < u[i] < Inf ? min(rtol * (u[i] - l[i]), atol) : atol for i = 1:n]
A1 = Array{Bool}(n)
A2 = Array{Bool}(n)
t1 = @benchmark active!($A1, $x, $l, $u, $δ)
show(STDOUT, MIME"text/plain"(), t1); println()
t2 = @benchmark $A2 .= active_dot.($x, $l, $u, $δ)
show(STDOUT, MIME"text/plain"(), t2); println()
@assert all(A1 .== A2)
end
foo(10000)
The explicit @inbounds-instrumented loop is still about 10x faster:
BenchmarkTools.Trial:
memory estimate: 0 bytes
allocs estimate: 0
--------------
minimum time: 5.528 μs (0.00% GC)
median time: 6.125 μs (0.00% GC)
mean time: 6.545 μs (0.00% GC)
maximum time: 23.742 μs (0.00% GC)
--------------
samples: 10000
evals/sample: 6
BenchmarkTools.Trial:
memory estimate: 0 bytes
allocs estimate: 0
--------------
minimum time: 45.809 μs (0.00% GC)
median time: 49.291 μs (0.00% GC)
mean time: 52.758 μs (0.00% GC)
maximum time: 229.375 μs (0.00% GC)
--------------
samples: 10000
evals/sample: 1
Can active_dot be improved somehow to close the gap? Thanks!