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What is begin...end in Erlang used for?

I just stomped at a begin...end in Erlang's documentation (here), but it doesn't give some examples of how it is useful.

Looking here in StackOverflow I found two cases where people would be using begin...end, both in list comprehensions:

  • https://stackoverflow.com/a/5645116/979505
  • https://stackoverflow.com/a/5141263/979505

But I wonder if there are more of such uses.

Can anyone provide another scenario in which a begin...end is useful in Erlang?

Thanks

like image 752
Daniel Avatar asked Nov 21 '13 15:11

Daniel


4 Answers

Macros, for example:

-define(M(A, B),
    begin
        C = foo(),
        bar(A, B, C)
    end).
like image 69
P_A Avatar answered Nov 07 '22 21:11

P_A


To evaluate a catch (always the same idea to have multiple expression reduced to one)

Res = (catch
    begin
        C = foo(Bar),
        io:format("evaluation of C ok~n"),
        D = bar(A, B, C)
     end),
like image 37
Pascal Avatar answered Nov 07 '22 23:11

Pascal


As previous answerers mentioned, this construct is used whenever you need to have multiple expressions but only one is allowed.

However, the majority of such cases would be considered a stinky style. I can remember only a few of places where a single expression is expected: an argument in a function call, catch expression, case of, try of and list comprehension. All of them except for list comprehension shouldn't be used with begin end construct because the variables are leaking to the outer scope probably causing the subsequent bindings to become matches.

List comprehension expression is different because it is transformed to a separate function with its own scope and no variable introduced in begin end leaks to the outer scope.

like image 5
Dmitry Belyaev Avatar answered Nov 07 '22 21:11

Dmitry Belyaev


According to erlang documentation is it block expression that evaluates each expression but returns only the last one.

See this example (not using block expression):

A = 1,
case A + 1 of
    3 ->
        ok;
    _->
        nop
end.

% returns ok

Now you can define A within the case argument using block expression:

case begin A = 1, A + 1 end of
    3 ->
        ok;
    _->
        nop
end.

%returns ok

That evaluates A = 1, then returns the result of A + 1.

Now we know that this will not work:

case A = 1, A + 1 of
    3 ->
        ok;
    _->
        nop
end.

% returns syntax error before: ','
like image 2
Aus Avatar answered Nov 07 '22 23:11

Aus