lud
Hi,
I’m pretty sure it is the case but I found conflicting informations online.
Can you confirm that the tail call optimization will take place for a function when only the first call is wrapped in a try/catch?
For instance this module:
defmodule T do
def top do
until(1_000_000_000)
catch
e -> {:got, e}
end
defp until(n, n) do
throw(:hello)
end
defp until(n, x) do
until(n, x + 1)
end
defp until(n) do
until(n, 0)
end
end
IO.inspect(T.top())
As you can see, the first call to until is wrapped in a try/catch but afterwards, there are no more wrapping blocks around the recursion. So TCO should be there, right?
I’m pretty sure this example would blow up if that were not the case anyway ![]()
Other case, with this code:
defmodule T do
def top do
until(1_000)
catch
e -> {:got, e}
end
defp until(n, m) when m >= n do
throw(:hello)
end
defp until(n, x) do
add =
try do
some_other_fun()
catch
_ -> 1
end
until(n, x + add)
end
defp until(n) do
until(n, 0)
end
defp some_other_fun do
case Enum.random(1..2) do
1 -> throw(:nope)
2 -> 2
end
end
end
The try/catch around some_other_fun does not wrap the tail call, so TCO is still there right?
Thanks!
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noelHan
I add one line to your code and get a result below
It looks like TCO is take place. You can try this with your second one
lud
Ah yes, indeed, why did I not think of it
adamu
I’m curious what you’re doing that requires
throw?lud
A parser and interpreter for user-defined scripts.