Simple Timeout of Function

General Usage
2

After digging around in various forums and trying various approaches, I’m very happy to share what appears to be a relatively general-purpose solution. The solution was provided by hhaensel here:

They define a macro as follows,

macro timeout(seconds, expr, fail)
    quote
        tsk = @task $expr
        schedule(tsk)
        Timer($seconds) do timer
            istaskdone(tsk) || Base.throwto(tsk, InterruptException())
        end
        try
            fetch(tsk)
        catch _
            $fail
        end
    end
end

They give the trivial application, which assigns x = 1 if the expression within begin/end takes less than 1 second, else it assigns x = “failed”.

x = @timeout 1 begin
    sleep(1.1)
    println("done")
    1
end "failed"

Playing around with this macro, it appears to work quite well when calling more complex libraries. For example, here are applications to NLsolve.jl and DifferentialEquations.jl:

using NLsolve

function f!(F, x)
        sleep(0.1)
    F[1] = (x[1]+3)*(x[2]^3-7)+18
    F[2] = sin(x[2]*exp(x[1])-1)
end

function j!(J, x)
    J[1, 1] = x[2]^3-7
    J[1, 2] = 3*x[2]^2*(x[1]+3)
    u = exp(x[1])*cos(x[2]*exp(x[1])-1)
    J[2, 1] = x[2]*u
    J[2, 2] = u
end

MaxTime = 10
res1 = @timeout MaxTime begin
    nlsolve(f!, j!, [ 0.1; 1.2]).zero
end NaN
println("Given MaxTime = 10 seconds, we get res1 = ", res1)

MaxTime = 0.1
res2 = @timeout MaxTime begin
    nlsolve(f!, j!, [ 0.1; 1.2]).zero
end NaN
println("Given MaxTime = 0.1 seconds, we get res2 = ", res2)

On my computer, this returns the correct solution when MaxTime = 10s, but returns NaN when MaxTime = 0.1s. Similarly for DifferentialEquations.jl we have,

using DifferentialEquations
function f(u, p, t) 
    sleep(0.001)
    return 1.01 * u
end
u0 = 1 / 2
tspan = (0.0, 1.0)
prob = ODEProblem(f, u0, tspan)
@time sol = solve(prob, Tsit5(), reltol = 1e-8, abstol = 1e-8)

MaxTime = 100
sol = @timeout MaxTime begin
    sol = solve(prob, Tsit5(), reltol = 1e-8, abstol = 1e-8)
end NaN
println("Given 100 seconds, sol converges, so sol(0.1) = ", sol(0.1))

MaxTime = 0.01
sol = @timeout MaxTime begin
    sol = solve(prob, Tsit5(), reltol = 1e-8, abstol = 1e-8)
end NaN

println("Given 0.01 seconds, sol doesn't converge, and sol = ", sol)

There are a few challenges. For example, BlackBoxOptim automatically recovers and returns a partial solution in case of an InterruptException, which is what the macro throws. You need to turn this off explicitly to make the macro work as intended:

using BlackBoxOptim

function rosenbrock2d(x)
    sleep(0.0001)
    return (1.0 - x[1])^2 + 100.0 * (x[2] - x[1]^2)^2
end

MaxTime = 1
x = @timeout MaxTime begin
    sleep(0.9)
    res = bboptimize(rosenbrock2d; SearchRange = (-5.0, 5.0), NumDimensions = 2, 
        TraceMode = :silent, RecoverResults=false)
    res
end NaN

println("Given MaxTime = 1 second, we get x = ", x)

MaxTime = 30
x = @timeout MaxTime begin
    sleep(0.9)
    res = bboptimize(rosenbrock2d; SearchRange = (-5.0, 5.0), NumDimensions = 2, 
        TraceMode = :silent, RecoverResults=false)
    res
end NaN

println("Given MaxTime = 30 seconds, we converge, with get best_fitness(x) = ", best_fitness(x))

I don’t know enough about the @task macro to determine if/when this might fail to work as intended, but for now it certainly seems a useful macro.

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