Erlang Efficiency quesitons

Robert Virding <>
Wed Mar 14 23:24:45 CET 2001

Francesco Cesarini <> writes:
>Let us put one together. Here are some more hints:
>* Use binaries when shuffling around large amounts of data between
>* Some one on the list had mentioned that appending binaries results in
>them being copied, thus avoid it.
>* In the documentation for the 4.4 extensions I found a note that 
>  f(X) when X == #rec{x=1, y=a} -> ... is less efficient than
> f(#rec{x=1, y=a} = X) -> ... 
>I assume because in A X is bound before the guard test, in B, the
>pattern match fails as soon as one of the conditions does not match.
>Back in the good old days, the recommendation was to use guards in Beam,
>while with the Jam there was no difference... 

A is inefficient because you build a new record and then do an equals test on 
the two structures.  It is also EXTREMELY unsafe as you get the default values 
on all unmentioned fields!

B just test the type of term and values of the given fields.  It is infact 
more efficient to do:

f(#rec{x=X,a=A}=Rec) ->


f(Rec) when record(Rec, rec) ->
    Rec#rec.x  Rec#rec.a

and it is more efficient to just test for a rec record with the first case 
than with the second.


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