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< Erlang ~ Erlang bit syntax: matching a NULL byte |
| parijat |
 Posted: Sun Aug 19, 2007 12:41 pm |
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Joined: 19 Aug 2007
Posts: 4
Location: Singapore
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Newbie learning Erlang. This is not exactly a problem; more like an observation.
The Erlang bit syntax is wonderful, if complicated.
Say you have a Binary like:
Code:
84> Bin1 = <<1,0,2>>.
<<1,0,2>>
Then, this will match:
Code: 85> <<A:1/binary, B:8, C:1/binary>> = Bin1.
<<1,0,2>> %% A => 1, C => 2
But this will not match:
Code: 86> <<A:1/binary, B:1/binary, C:1/binary>> = Bin1.
=ERROR REPORT==== 19-Aug-2007::17:46:17 ===
Error in process <0.139.0> with exit value: {{badmatch,<<3>>},[{erl_eval,expr,3}]}
** exited: {{badmatch,<<1,0,2>>,[{erl_eval,expr,3}]} **
It seems that I can specify a literal '0' as an int type (the default type) but not a binary type (see http://erlang.org/doc/reference_manual/expressions.html#bit_syntax for the gory details). Wonder why? One day I will read enough of the reference manual to understand. |
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| khigia |
| Posted: Mon Aug 20, 2007 1:47 am |
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Joined: 27 Mar 2007
Posts: 52
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Looking at the code, seems you entered all the expressions in a single erlang shell ... and the error on input 86 is probably not a binary matching problem.
In fact after the fisrt match line 85, variables A,B,C have values and types, with B being an integer of value 0. When matching again line 86, A and C match the binary but B do not match because the pattern line 86 is binary where B is already defined as an integer.
Note: in the shell, the function f/1 enable to "forget" the definition of a variable. |
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| parijat |
| Posted: Mon Aug 20, 2007 7:21 am |
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Joined: 19 Aug 2007
Posts: 4
Location: Singapore
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khigia wrote:
In fact after the fisrt match line 85, variables A,B,C have values and types, with B being an integer of value 0. When matching again line 86, A and C match the binary but B do not match because the pattern line 86 is binary where B is already defined as an integer.
Oh, yes. That is correct. Indeed, doing this works:
Code:
1> Bin = <<1, 0, 2>>.
<<1,0,2>>
2> <<A1:1/binary, B1:1/binary, C1:1/binary>>.
** 1: variable 'A1' is unbound **
3> <<A1:1/binary, B1:1/binary, C1:1/binary>> = Bin.
<<1,0,2>>
4> {A1, B1, C1}.
{<<1>>,<<0>>,<<2>>}
5> <<A2:1/binary, B2:8, C2:1/binary>> = Bin.
<<1,0,2>>
6> {A2, B2, C2}.
{<<1>>,0,<<2>>}
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| parijat |
| Posted: Mon Aug 20, 2007 3:10 pm |
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Joined: 19 Aug 2007
Posts: 4
Location: Singapore
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Hmm.. the problem that I originally faced was this:
Code:
(emacs@SIEGFRIED)1> Bin = <<1,0,2>>.
<<1,0,2>>
(emacs@SIEGFRIED)2> <<1:1/binary, 0:1/binary, 2:1/binary>> = Bin.
=ERROR REPORT==== 20-Aug-2007::20:35:47 ===
Error in process <0.35.0> on node 'emacs@SIEGFRIED' with exit value: {{badmatch,<<3 bytes>>},[{erl_eval,expr,3}]}
** exited: {{badmatch,<<1,0,2>>},[{erl_eval,expr,3}]} **
(emacs@SIEGFRIED)3> <<1:8, 0:8, 2:8>> = Bin.
<<1,0,2>>
That is, 0:1/binary does not match a null byte, but 0:8 does.
Sorry for getting lost while trying to "simplify" the problem. |
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| khigia |
| Posted: Tue Aug 21, 2007 12:28 am |
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Joined: 27 Mar 2007
Posts: 52
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I think it is the expected behaviour.
Line 3, it is possible to match the first byte of Bin to an integer of value 1, the second byte to a integer of value 0 and the third to integer 2.
But line 2, it is not possible to match the first byte of Bin to a 8bits binary of value "integer 1", the second byte of Bin to a 8bits binary of value "integer 0" etc
Matching <<A:1/binary, B:1/binary, C:1/binary>> = Bin
result in A = <<1>>, B = <<0>> and C = <<2>>,
not in A=1, B=0, C=2. |
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