Hi all.
With this code:
memory_bitmap.scan /\x00{5}+/
I get this warning:
test.rb:1: warning: ambiguous first argument; put parentheses or even
spaces
Could somebody please help me to understand what exactly is ambiguous
there?
Thank you!
-r
Hi all.
With this code:
memory_bitmap.scan /\x00{5}+/
I get this warning:
test.rb:1: warning: ambiguous first argument; put parentheses or even
spaces
Could somebody please help me to understand what exactly is ambiguous
there?
Thank you!
-r
On Aug 18, 2010, at 17:39 , Roger P. wrote:
Could somebody please help me to understand what exactly is ambiguous
there?
2 / 2
2 / 2 /
Roger P. wrote:
Hi all.
With this code:memory_bitmap.scan /\x00{5}+/
I get this warning:
test.rb:1: warning: ambiguous first argument; put parentheses or even
spacesCould somebody please help me to understand what exactly is ambiguous
there?
Thank you!
-r
The ambiguity comes from the double quantifier {5}+ where:
Given the string:
bitmap="\x30\x50-\x00\x00\x00\x00\x00\x00-\x40\x50\x30"
p bitmap.scan(/\x00+/) #=> ["\000\000\000\000\000\000"]
p bitmap.scan(/\x00{2}/) #=> ["\000\000", “\000\000”, “\000\000”]
HTH gfb
On Aug 19, 2010, at 11:26 AM, Gianfranco Bozzetti wrote:
bitmap=“\x30\x50-\x00\x00\x00\x00\x00\x00-\x40\x50\x30”
p bitmap.scan(/\x00+/) #=> [“\000\000\000\000\000\000”]
p bitmap.scan(/\x00{2}/) #=> [“\000\000”, “\000\000”, “\000\000”]HTH gfb
Posted via http://www.ruby-forum.com/.
I think the ambiguity is actually in parsing:
memory_bitmap.scan / \x00 { 5 } + /
Is that perhaps dividing by the result of calling \x00 with a block?
Spaces won’t help in this case, but some nice parentheses will do.
memory_bitmap.scan(/\x00{5}+/)
That will ensure that the parser sees /\x00{5}+/ as an argument to the
scan method. Whether the Regexp does what you want/expect is another
question.
-Rob
Rob B.
[email protected] http://AgileConsultingLLC.com/
[email protected] http://GaslightSoftware.com/
I think the ambiguity is actually in parsing:
memory_bitmap.scan / \x00 { 5 } + /
Is that perhaps dividing by the result of calling \x00 with a block?
Interesting. Ok thanks for all the responses.
It does work as expected
-r
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