crisefd

crisefd

Given

f = fn -> 2 end
g = fn -> 2 end

How can I match the result of g with f’s ?
If I do it like this, it fails because function calls are not allowed in pattern matching.

case g.() do
  f.() -> 1
  4 -> 2,
   5 -> 3,
  _ -> 6
end

Note:
This is dummy version of what I need to do in my project. Basically I need to pattern match the result of a function with the result of other functions. But don’t know how to go about it.

Showing Posts 1 to 9

kokolegorille

kokolegorille

What about this?

case {f.(), g.()} do
  {x, y} -> #whatever
end
benwilson512

benwilson512

Author of Craft GraphQL APIs in Elixir with Absinthe
f_result = f.()

case g.() do
  ^f_result -> 1
  4 -> 2,
   5 -> 3,
  _ -> 6
end
opsb

opsb

Given that you can just assign the result of a function call to a variable and match on that it’s always seemed surprising to me that the compiler doesn’t do this for you. Is there a technical reason for this not being supported or is it something that might happen at some point?

OvermindDL1

OvermindDL1

Matches are run in linear time, they can’t allow for arbitrary function calling or anything of the sort as they get compiled to a fairly efficient jumplist (sometimes).

NobbZ

NobbZ

a = fast_op()
b = long_running_thing()

case c do
  ^a -> 1
  ^b -> 2
end

Is very explicit that no matching would happen before both calls have returned, whereas the “sugared” version you suggest might hide this fact and a naive reader would assume that long_running_thing() never gets called when cmatches the result of fast_op().

OvermindDL1

OvermindDL1

And of course you’d generally want to run this one as this anyway:

cond do
  ^c = fast_op() -> 1
  ^c = long_running_thing() -> 2
end

^.^

NobbZ

NobbZ

Doesn’t this crash as soon as fast_op/0s return value isn’t matching cs value?

I think its this you mean:

cond do
  c === fast_op() -> 1
  c === long_running_thing() -> 2
end
OvermindDL1

OvermindDL1

Er, right, my bad, I meant this:

cond do
  match?(^c, fast_op()) -> 1
  match?(^c, long_running_thing()) -> 2
end

^.^;

Although using == would probably be better in this specific case, since it’s not a complex match at all:

cond do
  c == fast_op() -> 1
  c == long_running_thing() -> 2
end
silverdr

silverdr

That’s exactly what I was looking for as a replacement for the

which keeps getting over-verbose especially when more than two options are at play :slight_smile:

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