Confused about using Try in an enumerable

OK, bear with me with this silly example, but I’m still trying to get my head around all this stuff.

Suppose I have the following (admittedly stupid) situation: I have a Func<int, float> that might throw an exception, for example…

Func<int, float> f = n => 1 / n;

Obviously, if you pass zero into this function it will throw an exception. Suppose I also have an Enumerable<int> and want to apply the Func to it…

var result = nums.Select(n => f(n));

I guess I need to use Try to handle the fact that f might throw an exception…

var result = nums.Select(n => Try(() => f(n));

This gives me an IEnumerable<Try<float>>. What do I do with this now? Let’s say I have two different requirements (probably not at the same time)…

1. End up with an IEnumerable<float> that includes only the results that didn’t throw an exception
2. End up with (something) that includes the exceptions. I guess this would be an IEnumerable<Exception> but am not sure

Anyone able to explain how I would achieve each of these goals? Thanks

Again, please forgive the silly examples, but I find playing around like this helps me understand what’s going on.

[colethecoder]

In C# a double divided by zero is Infinity, so you wouldn’t get any exceptions, it’s nothing to do with LanguageExt’s Try.

If you go back to your original function n => 1 / n you should see the exceptions.

You can also avoid the ToEither bit in the example as there are extensions to support IEnumerable<Try<T>> so you can simply do:

IEnumerable<Try<int>> result = Range(0, 10)
                .Map(n => Try(() => 1 / n));
IEnumerable<int> r1 = result.Succs();
IEnumerable<Exception> r2 = result.Fails();

[MrYossu]

@colethecoder Thanks, that’s excellent!