Misalignment of text when Notepad++ edited files are opened in other editors (notepad, wordpad, google drive anyfile editor, etc.)
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Sure, here’s one of the 8086 assembly source files:
https://www.dropbox.com/s/gaxnn56hteuz5n7/g_lib.asm?dl=0
And here’s a screenshot of bad text alignment in the file. The comment block is shifted beyond 78 columns.
It just looks bad, and makes it a little awkward to read in a DOS editor.https://www.dropbox.com/s/b1b120zj5w28la2/bad_formatting2.png?dl=0
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Looks like your mixing tabs and spaces.
Activate the Show all characters
Convert it by using Edit->Blank Operations->TabToSpace !??
Cheers
Claudia -
Problem solved. Thanks!
Solution:
Settings->preferences->Language
check “Replace by space” boxEdit->Blank Operations->TAB to space
Make some arbitrary change to the file (type a character then delete it), then re-save the file.
Formatting/indentation should be fixed.
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@Compu-chan said:
So in your original post you provided misinformation:
Tabs have been replaced by spaces
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No, I checked the replace tab with spaces box before posting my question.
Someone else on a different forum mentioned it as a possible solution to a similar but different problem.It only prevents future tabulation symbols from appearing in your file.
It doesn’t get rid of the tabulations that are already present.You do that with:
Edit->Blank Operations->TAB to space -
Perhaps the verbage “Replace by space” in the Preferences is poor–it makes it sound like by checking the box it will perform an action rather than simply changing a setting!
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Also, i’m not sure if there’s already a way to do this, but being able to perfom operations on all open files at the same time would be nice.
I had to do the TAB to spaces operation on about a dozen files one by one to fix them all.
Not a big deal, but slightly inconvenient.Not sure if i’m supposed to mark this post as solved.
Don’t really see the option to do so. -
There is no option to do that specific operation on a group of files, however, it could be done with a regular expression Replace in Files on a bunch of files at once. Replace
\twith how many ever spaces you want a tab character to be.If you had changed the mode of this thread to “Ask as a Question” then you could later go in and “Mark as Solved”. You can still do it I think…
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Hi @compu-chan, @scott-sumner and All,
Scott said :
There is no option to do that specific operation on a group of files, however, it could be done with a regular expression Replace in Files on a bunch of files at once. Replace
\twith how many ever spaces you want a tab character to be.Unfortunately, Scott-, it’s not that simple !! Just because the physical length of the tabulation character depends on its position :-(( So I began to investigate a way to simulate all the native Blank Operations of Notepad++, by a regex search replacement.
The main advantage, of this method, is that you can perform these Blank operations, on multiple files, belonging to a same folder, in the Find in files dialog :-))
The first 3 operations are quite easy to realize :
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Trim Trailing Space :
- SEARCH
[\x20\t]+$and REPLACElet the zone EMPTY
- SEARCH
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Trim Leading Space :
- SEARCH
^[\x20\t]+and REPLACElet the zone EMPTY
- SEARCH
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Trim Leading or Trailing Space :
- SEARCH
^[\x20\t]+|[\x20\t]+$and REPLACELet the zone EMPTY
- SEARCH
The following 2 operations are not too difficult to achieve, too !
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EOL to Space :
- SEARCH
(?<=\x20)\R|(\R)and REPLACE(?1\x20)( Change of any line-break by a space or suppression, if preceded by a space character )
- SEARCH
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Remove Unnecessary Blank and EOL :
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SEARCH
^[\x20\t]+|[\x20\t]+$and REPLACELet the zone EMPTY( Suppression of leading and trailing Blank characters ) -
SEARCH
\Rand REPLACE\x20( Replacement of any line break by a space character )
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Now, due to the tabulation’s behaviour, which always stops at column
4*nwhatevern > 0, the last 3 Blank Operations are much hardier to elaborate, on the “regex” point of view !!-
TAB to Space :
- SEARCH
(?-s)(?:()|(.)|(..)|(...))\t|(....)and REPLACE(?1 )(?2\2 )(?3\3 )(?4\4 )(?5$0)
- SEARCH
Briefly, here are, below, the different cases :
Let C = Any unique STANDARD character < = () + \t -> \1 + 4 Spaces Group 1 1 < = (C) + \t -> \2 + 3 Spaces Group 2 < = (C) + \t -> \2 + 3 Spaces Group 2 12 < = (CC) + \t -> \3 + 2 Spaces Group 3 < = (CC) + \t -> \3 + 2 Spaces Group 3 123 < = (CCC) + \t -> \4 + 1 Space Group 4 < = (CCC) + \t -> \4 + 1 Space Group 4 1234< = (CCCC) -> $0 Group 5Notes :
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Depending on the number of characters, preceding the tabulation character, this S/R rewrites these characters, followed by the appropriate number of spaces
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Note that if the range of 4 chars does not contain any tabulation, it is simply rewritten (
$0). This replacement, seemingly useless, is, however, necessary to go on looking for the next blocks of 4 positions long !
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Space to TAB (All) :
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SEARCH
(?-s)(?|([^ \t\r\n])\x20(?:\x20[\x20\t]|\t)|([^ \t\r\n]{2})\x20[\x20\t]|([^ \t\r\n]{3})\x20)|(\x20{0,3}\t|\x20{4})|([^ \t\r\n]{1,3}\t|....) -
REPLACE
(?1\1\t)(?2\t)(?3$0)
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Again, below, here is the recapitulation of all cases, with their appropriate replacements :
Let C = [^ \t\r\n] = Any unique STANDARD character, different of a SPACE and a TABULATION a < = (C) + 1 sp + \t -> \1\t Group 1 a < = (C) + 2 sp + \t -> \1\t Group 1 a < = (C) + 3 sp -> \1\t Group 1 ab < = (CC) + 1 sp + \t -> \1\t Group 1 ab < = (CC) + 2 sp -> \1\t Group 1 abc < = (CCC) + 1 sp -> \1\t Group 1 < = (0 sp + \t) -> \t Group 2 < = (1 sp + \t) -> \t Group 2 < = (2 sp + \t) -> \t Group 2 < = (3 sp + \t) -> \t Group 2 < = (4 sp) -> \t Group 2 a < = (C + \t) -> $0 Group 3 ab < = (CC + \t) -> $0 Group 3 abc < = (CCC + \t) -> $0 Group 3 abcd< = (CCCC) -> $0 Group 3Notes :
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Quickly, all the cases are divide up into 3 main parts, with the appropriate replacements :
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Standard characters followed by a mix of spaces/tabulations must be rewritten, followed by a tabulation (
(?1\1\t)) -
Mix of spaces/tabulation, only, must be replaced by a single tabulation (
(?2\t)) -
Standard characters, followed by an unique tabulation, have to be simply rewritten (
(?3$0))
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Note, in the search regex, a special construct
(?!......), which resets the sub-expression count, at the start of each|alternative of this construct. So, whatever the branch matched, in our example, the matched expression is always stored in group 1 and will be replaced, according to the conditional form(?1\1\t)
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Space to TAB (Leading) :
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SEARCH
^(?:\x20|\t)+and REPLACE$0# -
SEARCH
(?:\x20{4}|\x20{0,3}\t)(?=.*#)#?and REPLACE\t
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Here are all the cases, of blank characters combination, which may occur, at beginning of lines :
< = 0 sp + \t -> \t < = 1 sp + \t -> \t < = 2 sp + \t -> \t < = 3 sp + \t -> \t < = 4 sp -> \tNotes :
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In the first S/R, we look, at beginning of lines, any range of space or tabulation character(s) and simply rewrite this range,
$0, followed with a#symbol. Note that you may use any symbol, absent from your file, which will be used as a mark -
In the second S/R, all the combination of leading blank characters, which are followed, further on, by the
#character, are changed, along with the mark character, with an unique tabulation character !
Best Regards,
guy038
PS :
Let consider the special branch reset construct, below :
(a)(?|x(y)z|(p(q)r)|(t)u(v))(z)We have :
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Group 1 =
a -
Group 2 =
yorp(q)rort -
Group 3 = None or
qorv -
Group 4 =
z
With a classical list of alternatives, as below :
(a)(?:x(y)z|(p(q)r)|(t)u(v))(z)We would get, instead :
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Group 1 =
a -
Group 2 = None or
y -
Group 3 = None or
p(q)r -
Group 4 = None or
q -
Group 5 = None or
t -
Group 6 = None or
v -
Group 7 =
z
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@guy038 said:
the physical length of the tabulation character depends on its position
When I said “Replace \t with how many ever spaces you want a tab character to be” I was considering leading tab characters only, which is really the only consistent way to use tab characters (IMO). I tend to stay away from tab characters in general, but when I work on projects where they are used, it is always in the “tab-indent, space-align” style. This means that tab characters are the only whitespace that are allowed before the first non-whitespace character on a line, and spaces are the only valid whitespace after the first non-whitespace character on a line. With this usage, my original Replace operation is valid–there should never be a situation in which moving to the next tab-stop is not the full # of spaces (if the conversion were to be done).
due to the tabulation’s behaviour, which always stops at column 4*n whatever n > 0
Can you explain this, I’m not understanding what this means??
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Hi, @scott-sumner,
Indeed, using leading tabulation characters and space characters, everywhere else, in lines, should be the sensible attitude, while coding :-D And, therefore, your solution, about leading tabs is quite exact !
As for my assertion :
due to the tabulation’s behaviour, which always stops at column
4*nwhatevern > 0I pointed out the fact that a tabulation character always ends at column 4, 8, 12, 16,…, that is to say, on a
4*nposition !Moreover, if
cis the column, > 0, where begins the tabulation character, its physical lengthl, between 1 and 4, can be found with the formulal = 4 - ((c-1) % 4)BTW, I just realize that all this works, only, if the tab size value, in Settings > Preferences… > Language > Tab Settings, is
4. In the general case, if tab size =s, the physical lengthlof a tabulation, beginning at columnc, would be :l = s - ((c-1) % s), with%standing for the mathematical operation modulo !Cheers,
guy038
P.S : I forgot to give an example :
Let’s suppose the tab size is
7and a tabulation character begins at column c =157. This implies that its length l = 7 - ((157-1) % 7) =5and it ends at column 157 + 5 - 1 =161, which is, effectively, a multiple of 7 ( 161 = 7 * 23 ) -
Thanks…what confused me was that it sounded like you were saying that tabstops were always 4 columns…your most-recent post clears that up.
