I'm trying to replace a code using Regex in visual studio community, but I'm not getting it.
ZeroMemory(&ab,15);
ZeroMemory(&abc,30);
ZeroMemory(&tickt,sizeof(tickt));
To replace
memset(&ab,0,15);
memset(&abc,0,30);
memset(&tickt,0,sizeof(tickt));
I need to replace zeromemory with memset and the first one with ,0,
thank you all in advance
Clearly you are asking about the "Find and Replace" feature in Visual Studio.
As a starting point, something like this should do what you want:
From: ZeroMemory\s*\((.*),
To: memset($1, 0,
Because parentheses are used in regular expressions for matching groups, you need to escape them (\() if you want to match an actual parenthesis.
Then we use (.*) to create a group that captures whatever your first parameter is. Note that this is very naive. If you have a non-trivial expression for that argument (e.g. a function call where a comma appears anywhere) this will likely fail.
So that matches the stuff you want, and then for the replacement you can use $1 to substitute the first matching group (in which we captured your first argument).
The only extra detail I added was a \s* between the function identifier and the parenthesis. This matches optional whitespace characters, so it will still match even if you have stuff like:
ZeroMemory (foo, bar);
I recommend you read the documentation to familiarize yourself with regular expressions in Visual Studio.
Related
I have a need to replace this:
fixed variable 123
with this:
fixed variable 234
In VSCode this matches fine:
fixed(.*)123
I can't find any way to make it put the capture in the output if a number follows:
fixed$1234
fixed${1}234
But the find replace window just looks like this:
I read that VSCode uses rust flavoured rexes.. Here indicates ${1}234 should work, but VSCode just puts it in the output..
Tried named capture in a style according to here
fixed(?P<n>.*)123 //"invalid regular expression" error
VSCode doesn't seem to understand ${1}:
ps; I appreciate I could hack it in the contrived example with
FIND: fixed (.*) 123
REPL: fixed $1 234
And this does work in vscode:
but not all my data consistently has the same character before the number
After a lot of investigation by myself and #Wiktor we discovered a workaround for this apparent bug in vscode's search (aka find across files) and replace functionality in the specific case where the replace would have a single capture group followed by digits, like
$1234 where the intent is to replace with capture group 1 $1 followed by 234 or any digits. But $1234 is the actual undesired replaced output.
[This works fine in the find/replace widget for the current file but not in the find/search across files.]
There are (at least) two workarounds. Using two consecutive groups, like $1$2234 works properly as does $1$`234 (or precede with the $backtick).
So you could create a sham capture group as in (.*?)()(\d{3}) where capture group 2 has nothing in it just to get 2 consecutive capture groups in the replace or
use your intial search regex (.*?)(\d{3}) and then use $` just before or after your "real" capture group $1.
OP has filed an issue https://github.com/microsoft/vscode/issues/102221
Oddly, I just discovered that replacing with a single digit like $11 works fine but as soon as you add two or more it fails, so $112 fails.
I'd like to share some more insights and my reasoning when I searched for a workaround.
Main workaround idea is using two consecutive backreferences in the replacement.
I tried all backreference syntax described at Replacement Strings Reference: Matched Text and Backreferences. It appeared that none of \g<1>, \g{1}, ${1}, $<1>, $+{1}, etc. work. However, there are some other backreferences, like $' (inserts the portion of the string that follows the matched substring) or $` (inserts the portion of the string that precedes the matched substring). However, these two backreferences do not work in VS Code file search and replace feature, they do not insert any text when used in the replacement pattern.
So, we may use $` or $' as empty placeholders in the replacement pattern.
Find What: fix(.*?)123
Replace With:
fix$'$1234
fix$`$1234
Or, as in my preliminary test, already provided in Mark's answer, a "technical" capturing group matching an empty string, (), can be introduced into the pattern so that a backreference to that group can be used as a "guard" before the subsequent "meaningful" backreference:
Find What: fixed()(.*)123 (see () in the pattern that can be referred to using $1)
Replace With: fixed$1$2234
Here, $1 is a "guard" placeholder allowing correct parsing of $2 backreference.
Side note about named capturing groups
Named capturing groups are supported, but you should use .NET/PCRE/Java named capturing group syntax, (?<name>...). Unfortunately, the none of the known named backreferences work replacement pattern. I tried $+{name} Boost/Perl syntax, $<name>, ${name}, none work.
Conclusion
So, there are several issues here that need to be addressed:
We need an unambiguous numbered backerence syntax (\g<1>, ${1}, or $<1>)
We need to make sure $' or $` work as expected or are parsed as literal text (same as $_ (used to include the entire input string in the replacement string) or $+ (used to insert the text matched by the highest-numbered capturing group that actually participated in the match) backreferences that are not recognized by Visual Studio Code file search and replace feature), current behavior when they do not insert any text is rather undefined
We need to introduce named backreference syntax (like \g<name> or ${name}).
I try to build simple regular expression to remove some parts of bad (unwanted) code and needed use look behind feature.
It worked until i added \s+ to it to exclude spaces from mark.
Eliminating parts of expression i finally got to (?<=\s+)foo which is still warned as invalid expression.
It may looks a little weird or unclear so expanding it:
(?<=foo\s+)bar is warned as invalid expression, where (?<=foo)\s+bar is working but it marks spaces before bar.
I am use it in notepad++.
From the comment by #Toto Notepad++ does not support variable length lookbehind. It uses the boost regex.
Notepad++ does support \K to reset the starting point of the reported match.
\bfoo\s+\Kbar\b
Regex demo
Another way is to capture bar in a capturing group.
\bfoo\s+(bar)\b
Regex demo
Note that \s also matches a newline, and perhaps you might also use \h+ to match 1+ horizontal whitespace characters.
I'd like to replace automatically all the strings in my solution which are like this one
NotifyPropertyChanged("VariableParameter")
with this
NotifyPropertyChanged(Function() VariableParameter)
by "Quick Replace" in "Find and Replace" using a regular expression in Visual Studio 2010.
I have not the slightest idea how to do this when I have to keep each different variable parameter.
Try the following pattern and replacement.
Pattern: NotifyPropertyChanged\("{[^"]+}"\)
This matches your text, while escaping the parentheses. The {[^"]+} portion tags the contents (via the curly braces) and the [^"]+ bit matches any character that isn't a double-quote, one or more times.
Replacement: NotifyPropertyChanged(Function() \1)
This replaces the matched text and is fairly straightforward to understand. The \1 portion refers to the first (and only, in this example) tagged text from the pattern, which is the content between double-quotes.
I need to convert some variables that use parentheses instead of bracket in a Javascript file. For example, MyVariable(i) should be MyVariable[i].
I use the Find and Replace tool of Visual Studio with this Regex :
MyVariable\({(.+)}\)
and replace with:
MyVariable\[\1\]
This work fine for case like :
asdas.MyVariable(i+1)
asdas.MyVariable(i)
asdas.MyVariable(i).asd
MyVariable(i+1)
But doesn't work for case like
if (parseInt(OtherObject.MyVariable(i+1).Dest.XYZ)==SINK_STATE_TYPE || OtherObject.MyVariable(i).X == "ok")
The last one will do take the first parentheses and the last one of the line and act weird. What do I need to change in the Regex to be able to make it works with line that has multiple parentheses.
Your pattern is greedy. The (.+) part matches everything, including a closing parenthesis ) character until it reaches the last ) character. To make it non-greedy you can use # instead of +, which means "match one or more occurrences of the preceding expression, and match as few characters as possible." In .NET this is usually handled by placing a ? after it, such as .+? but the Visual Studio regex flavor is different and uses the # metacharacter instead.
The updated pattern becomes:
MyVariable\({(.#)}\)
You can also use [^)]+, which matches any character that is not a ) character. When it encounters a ) it will stop matching. This alternate pattern is:
MyVariable\({([^)]+)}\)
If you're familiar with the .NET regex flavor you might be interested in checking out the Visual Studio 2010 Productivity Power Tools, which has extended Find/Replace to allow the .NET regex flavor to be used.
I'm in the process of converting some LaTeX documentation to restructured text and having some trouble with a regular expression in Visual Studio 2003. I'm trying to convert \emph{text} to *text* using the following find/replace strings:
\\emph\{([^\}]*)\}
*\0*
However, using this pair I get \emph{text} converted to *\emph{text}* which was not what I expected. When I use *\1* instead of *\0* I get ** as the replacement result.
What am I missing or what don't I understand about the grouping rules?
Thanks.
I think that in the VS regex replacement syntax, \0 is the whole matched string, while \1 is the content of the first captured variable (\2 being the second and so on). Therefore:
\0
However, using this pair I get
\emph{text} converted to *\emph{text}*
which was not what I expected.
Thus confirming, \0 is the whole matched string.
When I use *\1* instead of *\0* I get ** as the replacement result.
Probably you are not matching anything in the capture class.
To add more detail, the syntax for defined a capture class (called a tagged expression in the docs) uses the braces {}, not parentheses () as you are using here. Probably this will work as the "find" expression:
\\emph\{{[^\}]*}\}