Julia: Overflow in Timer when passing time as an Int

[KristofferC]

I was trying to make a message be sent 18446744073709552 seconds (585 million years) into the future by doing the following:

julia> Timer(18446744073709552) do timer
          println("Hello distant friend")
       end;

However, the message prints immediately and my distant friend will not receive the greeting. :(

[ararslan]

I think something like this might help:

--- a/base/asyncevent.jl
+++ b/base/asyncevent.jl
@@ -71,8 +71,8 @@ mutable struct Timer
     function Timer(timeout::Real; interval::Real = 0.0)
         timeout ≥ 0 || throw(ArgumentError("timer cannot have negative timeout of $timeout seconds"))
         interval ≥ 0 || throw(ArgumentError("timer cannot have negative repeat interval of $interval seconds"))
-        timeout = UInt64(round(timeout * 1000)) + 1
-        interval = UInt64(round(interval * 1000))
+        timeout = round(UInt64, timeout) * 1000 + 1
+        interval = round(UInt64, interval) * 1000
         loop = eventloop()

         this = new(Libc.malloc(_sizeof_uv_timer), ThreadSynchronizer(), true, false)

[KristofferC]

Hm, why would that help? Seems the * 1000 can still easily overflow.

[ararslan]

When timeout is signed, it doesn't have as much room in the upper range before overflow, so if we make it unsigned first, there's a bit more room to make it bigger before it overflows.

[KristofferC]

Yeah, a factor of 2. The easiest to me seems to just error if it overflows.

[vtjnash]

The easiest to me seems to just error if it overflows

Or just don't worry about starting the Timer if you ask for more than 1 millennium into the future

[KristofferC]

The problem is someone that puts a very long default time for the timer to keep the same code and it then fires early. I don't think all olqcew that start a timer should need to worry about overflow. Seems simple enough to fix.