Erlang generate anonymous function of an arbitary arity?

user_1245380 0
Viewed 94

Is it possible to write a function that returns an anonymous function of a specified arity? I'd like to be able to generate a function that can be passed to meck:expect/3 as the third argument so I can dynamically mock existing functions of any arity?

I've done quite a bit of searching and it seems like the only way to solve this is by hardcoding things like this:

gen_fun(1, Function) ->
    fun(A) -> Function([A]) end;
gen_fun(2, Function) ->
    fun(A, B) -> Function([A, B]) end;
...
arity erlang
1 Answers

It's not pretty, but you can use the same trick as the shell and build your functions from the ground up:

-module(funny).

-export([gen_fun/1, gen_fun/2]).

-spec gen_fun(function()) -> function().
gen_fun(Function) ->
    {arity, Arity} = erlang:fun_info(Function, arity),
    gen_fun(Arity, Function).

-spec gen_fun(non_neg_integer(), function()) -> function().
gen_fun(Arity, Function) ->
    Params = [{var, 1, list_to_atom([$X| integer_to_list(I)])} || I <- lists:seq(1, Arity)],
    Anno = erl_anno:new(1),
    Expr =
        {'fun',
         Anno,
         {clauses, [{clause, Anno, Params, [], [{call, Anno, {var, Anno, 'Function'}, Params}]}]}},
    {value, Fun, _Vars} = erl_eval:expr(Expr, [{'Function', Function}]),
    Fun.

Then, in the shell…

1> F = funny:gen_fun(fun io:format/2).
#Fun<erl_eval.43.40011524>
2> F("~ts~n", [""]).

ok
3> F1 = funny:gen_fun(fun io:format/1).
#Fun<erl_eval.44.40011524>
4> F1("~n").

ok
5>
user_28818650
Related