Nul characters and wildcards in regexp

1 Ansicht (letzte 30 Tage)
Brig
Brig am 23 Mär. 2015
It seems to be the case that when a regular expression encounters the pattern
[0 46 0]
it always matches everything after it.
Is this a side effect explainable in any way, or just a bug?
Example:
K>> regexp(char([0 0 1 0 41 41 41 41 41 41]),char([0 '.' 0 40 40 40 40]))
ans =
2
The expected answer would be [] because 41 should not match the 40 in the pattern.
  1 Kommentar
per isakson
per isakson am 23 Mär. 2015
Bearbeitet: per isakson am 23 Mär. 2015
I call it a bug. Why not report it to tech-support whether it is a real problem to your projekt or not.

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Antworten (1)

Tijs Van Oevelen
Tijs Van Oevelen am 10 Mai 2016
Hi,
this bug is probably already fixed. I tested your example in several versions of Matlab:
  • in R2013b:
>> regexp(char([0 0 1 0 41 41 41 41 41 41]),char([0 '.' 0 40 40 40 40]))
ans =
2
  • in R2015a:
>> regexp(char([0 0 1 0 41 41 41 41 41 41]),char([0 '.' 0 40 40 40 40]))
ans =
2
  • in R2016a:
>> regexp(char([0 0 1 0 41 41 41 41 41 41]),char([0 '.' 0 40 40 40 40]))
ans =
[]

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