Where is error message on the remote worker?

General Usage
,
3

I understand that remotecall_fetch will return the error. However, the function that is going to be executed on the remote process will never return and will always be waiting for tasks from the master process:

mutable struct ParallelFunction
    jobs::RemoteChannel
    results::RemoteChannel
    nreceived::Int64
    ntaken::Int64
end


function do_work(f, jobs, results, args...)
    while true
        x = take!(jobs)
        result = f(x, args...)
        put!(results, (x, result))
    end
end


function ParallelFunction(f, args...; len=128)
    jobs = RemoteChannel(()->Channel{Any}(len))
    results = RemoteChannel(()->Channel{Any}(len))
    for p in workers()
        remote_do(do_work, p, f, jobs, results, args...)
    end
    return ParallelFunction(jobs, results, 0, 0)
end


function give!(pf::ParallelFunction, x)
    put!(pf.jobs, x)
    pf.nreceived += 1
end


function get!(pf::ParallelFunction)
    result = take!(pf.results)
    if result[2] isa ErrorException
        throw(result[2])
    end
    pf.ntaken += 1
    return result
end

Suppose a function f has four arguments x, a, b and c. The purpose is to evaluate the function f at several different xs with fixed argument a, b and c.
A ParallelFunction can be constructed as

pf = ParallelFunction(f, a, b, c; len=128)

To evaluate the function f at x1, x2 and x3, the following statement can be used

give!(pf, x1)
give!(pf, x2)
give!(pf, x3)

The function will be evaluated in parallel on any possible processes. To obtain the result:

for i in 1:3
    println(get!(pf))
end

get!(pf) returns a tuple, the first element of the tuple is the argument x and the second argument is the corresponding result.

However, it is possible that the user forget to define the function f on the remote process. In this case, the get!(pf) will hang indefinitely, which is not a desirable result. I want to throw an error if this is the case.