[erlang-questions] binary match vs. is_binary in optimization

[Alex S.]

t1 doesn't extract a sub-binary from your binary, so there's nothing to
optimize.

2017-08-02 19:35 GMT+03:00 Andreas Schultz <aschultz@REDACTED>:

> Hi,
>
> Can someone explain why the binary optimization treats a is_binary()
> guard different that an binary match?
>
> I the attached test code, I get a:
>
> > test.erl:7: Warning: NOT OPTIMIZED: called function t1/1 does not begin
> with a suitable binary matching instruction
>
> and a
>
> > test.erl:14: Warning: OPTIMIZED: creation of sub binary delayed
>
> for the second function. IMHO both should be identical, but apparently
> they are not.
>
> Many thanks,
> Andreas
>
> Test code:
>
> -module(test).
>
> -compile([bin_opt_info]).
>
> -export([do/1]).
>
> do(<<Bin/binary>>) ->
>     t1(Bin),
>     t3(Bin).
>
> t1(Bin) when is_binary(Bin) ->
>     test(Bin, []).
>
> t3(<<Bin/binary>>) ->
>     test(Bin, []).
>
> test(<<X:4/integer, _/binary>>, _) ->
>     X.
> --
> Dipl.-Inform. Andreas Schultz
>
> email: as@REDACTED
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