Can someone explain to me how parameter pattern matching works exactly? def foo({a,b}) do ... end foo({1,2}) Pattern matching that happens: {a,b} = {1,2} def equal(a,a), do: true equal(1,1) Pattern matching that happens? I was wondering because if it happens like a=1 and then a=1 then equal(1,…

During compilation def equal(a,a), do: true is basically rewritten to def equal(a,b) when a === b, do: true Both these functions compile to the exact same bytecode.

a=1 and then a=2 is totally legal It is not actually. The assignment happens once (a=1), and a=2 would mean 1=2 which doesn’t match.

It’s explained here: https://elixir-lang.org/getting-started/pattern-matching.html If a variable is mentioned more than once in a pattern, all references should bind to the same pattern: iex> {x, x} = {1, 1} {1, 1} iex> {x, x} = {1, 2} ** (MatchError) no match of right hand side value: {1, 2}

So am i right to say that for parameter pattern matching, the parameters are taken in as a tuple and tuple pattern matching happens def equal(a,a), do: true equal(1,1) Patter n matching that happens: {a,a} = {1,1} Hence {a,b} = {1,2} matching fails

Not really, because this happens even with a single parameter. def equal(%{x: a, y: a}), do: true will return true for %{x: 1, y: 1} and raise for %{x: 1, y: 2}.