Enuzo
Order of precedence in string matching
Hello there,
I know that:
str = "_forum"
"elixir" <> str
"elixir_forum"
But the following fails:
"elixir" <> str = "_forum"
** (MatchError) no match of right hand side value: "_forum"
And the following works:
iex(1)> "elixir" <> str = "elixir_forum"
"elixir_forum"
iex(2)> str
"_forum"
This is making my head spin. Please, can someone help me with a logical breakdown of how this matching, binding, and concatenation operation works and the rationale behind the ordering?
Most Liked
kip
ex_cldr Core Team
The key is to recognise that = is really a match operator, with the side affect of binding the variable.
Therefore:
str = "_forum"matches and binds. There is no pattern so anything will bind."elixir" <> strhas no pattern matching (there’s no left and right), its just a call to the functionKernel.<>/2with the parameters"elixir"andstr"elixir" <> str = "_forum"is a pattern match (there is a left and right around thematchoperator=. It can’t match because"_forum"cannot match"elixir" <> str"elixir" <> str = "elixir_forum"matches just fine as I think you may have worked out by now
Hope that’s sufficiently clear, if not ask away
7
kip
ex_cldr Core Team
Not really analagous, but but they are both matches.
"elixir" <> str = "elixir_forum"is a binary match (thats indicated by the<>). Basically it can be read as "match a string that starts withelixirand bind anything that comes after that tostr.strend up being bound to"_forum"["elixir", str] = ["elixir", "forum"]is asking the question: are both sides of the match operator (the=) of the same shape? In this case, yes - they are both lists with 2 elements. Then do the elements match? For the first element, yes,"elixir"=="elixir. And the second element binds"forum"tostr.
3
hauleth
"elixir" <> str = "elixir_forum"
Is exactly the same as:
<<"elixir", str::binary>> = "elixir_forum"
3
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