I look here ANSI C grammar .
This page includes a lot of regular expressions in Lex/Flex for ANSI C.
Having a problem in understanding regular expression for string literals.
They have mentioned regular expression as \"(\\.|[^\\"])*\"
As I can understand \" this is used for double quotes, \\ is for escape character, . is for any character except escape character and * is for zero or more times.
[^\\"] implies characters except \ , " .
So, in my opinion, regular expression should be \"(\\.)*\".
Can you give some strings where above regular expression will fail?
or
Why they have used [^\\"]?
The regex \"(\\.)*\" that you proposed matches strings that consist of \ symbols alternating with any characters like:
"\z\x\p\r"
This regular expression would therefore fail to match a string like:
"hello"
The string "hello" would be matched by the regex \".*\" but that would also match the string """" or "\" both of which are invalid.
To get rid of these invalid matches we can use \"[^\\"]*\", but this will now fail to match a string like "\a\a\a" which is a valid string.
As we saw \"(\\.)*\" does match this string, so all we need to do is combine these two to get \"(\\.|[^\\"])*\".
Related
I'm trying to create a regular expression match for an email address and I intend to use it in a dart application.
I found the following regex for that:
(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|"(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21\x23-\x5b\x5d-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])*")#(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\[(?:(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9]))\.){3}(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9])|[a-z0-9-]*[a-z0-9]:(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21-\x5a\x53-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])+)\])
now i'm really new to dart but I understood that I can create regular expression strings with r'' or r"".
now with dart I can escape characters with \ so if I want to escape an apostrophe in a string that I started and ended with apostrophe I can just do this:
final String a = 'foo\'bar';
but with final String a = r'foo\'bar' I get an error. how can I properly escape that ?
thank you
No, r'' does not mean "regular expression". It means "raw", so backslash is interpreted as a literal backslash, and not as an escape character.
Not having to escape each backslash is useful for the kind of strings which often contain a lot of backslashes, such as regular expression patterns.
Regular expressions are created as instances of the RegExp class.
You can concatenate raw strings that use different delimiters to create a single string for the whole pattern. In your case, this should work:
String pattern = r"(?:[a-z0-9!#$%&'*+/=?^_`{|}~-]+(?:\.[a-z0-9!#$%&'*+/=?^_`{|}~-]+)*|" + r'"(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21\x23-\x5b\x5d-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])*")#(?:(?:[a-z0-9](?:[a-z0-9-]*[a-z0-9])?\.)+[a-z0-9](?:[a-z0-9-]*[a-z0-9])?|\[(?:(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9]))\.){3}(?:(2(5[0-5]|[0-4][0-9])|1[0-9][0-9]|[1-9]?[0-9])|[a-z0-9-]*[a-z0-9]:(?:[\x01-\x08\x0b\x0c\x0e-\x1f\x21-\x5a\x53-\x7f]|\\[\x01-\x09\x0b\x0c\x0e-\x7f])+)\])';
RegExp exp = new RegExp(pattern);
I've got a little problem with regex.
I got few strings in one file looking like this:
TEST.SYSCOP01.D%%ODATE
TEST.SYSCOP02.D%%ODATE
TEST.SYSCOP03.D%%ODATE
...
What I need is to define correct regex and change those string name for:
TEST.D%%ODATE.SYSCOP.#01
TEST.D%%ODATE.SYSCOP.#02
TEST.D%%ODATE.SYSCOP.#03
Actually, I got my regex:
r".SYSCOP[0-9]{2}.D%%ODATE" - for finding this in file
But how should look like the changing regex? I need to have the numbers from a string at the and of new string name.
.D%%ODATE.SYSCOP.# - this is just string, no regex and It didn't work
Any idea?
Find: (SYSCOP)(\d+)\.(D%%ODATE)
Replace: $3.$1.#$2 or \3.\1.#\2 for Python
Demo
You may use capturing groups with backreferences in the replacement part:
s = re.sub(r'(\.SYSCOP)([0-9]{2})(\.D%%ODATE)', r'\3\1.#\2', s)
See the regex demo
Each \X in the replacement pattern refers to the Nth parentheses in the pattern, thus, you may rearrange the match value as per your needs.
Note that . must be escaped to match a literal dot.
Please mind the raw string literal, the r prefix before the string literals helps you avoid excessive backslashes. '\3\1.#\2' is not the same as r'\3\1.#\2', you may print the string literals and see for yourself. In short, inside raw string literals, string escape sequences like \a, \f, \n or \r are not recognized, and the backslash is treated as a literal backslash, just the one that is used to build regex escape sequences (note that r'\n' and '\n' both match a newline since the first one is a regex escape sequence matching a newline and the second is a literal LF symbol.)
I'm trying some string manipulation using regex's, but I'm not getting the expected output
var myString = "/api/<user_id:int>/"
myString.replace(Regex("<user_id:int>"), "(\\d+)")
this should give me something like /api/(\d+)/ but instead I get /api/(d+)/
However if I create an escaped string directly like var a = "\d+"
I get the correct output \d+ (that I can further use to create a regex Pattern)
is this due to the way String::replace works?
if so, isn't this a bug, why is it removing my escape sequences?
To make the replace a literal string, use:
myString.replace(Regex("<user_id:int>"), Regex.escapeReplacement("(\\d+)"))
For details, this is what kotlin Regex.replace is doing:
Pattern nativePattern = Pattern.compile("<user_id:int>");
String m = nativePattern.matcher("/api/<user_id:int>/").replaceAll("(\\d+)");
-> m = (d+)
From Matcher.replaceAll() javadoc:
Note that backslashes () and dollar signs ($) in the replacement
string may cause the results to be different than if it were being
treated as a literal replacement string. Dollar signs may be treated
as references to captured subsequences as described above, and
backslashes are used to escape literal characters in the replacement
string.
The call to Regex.escapeReplacement above does exactly that, turning (\\d+) to (\\\\d+)
You are using a .replace overload that takes a regex as the first argument, thus, the second argument is parsed as a regex replacement pattern. Inside a regex replacement pattern, a \ char is special, it may escape a dollar symbol to be treated as a literal dollar sign. So, the literal backslash inside regex replacement patterns should be doubled.
You might use
myString.replace(Regex("<user_id:int>"), """(\\d+)""")
Whenever you have to search and replace with a regex and your replacement pattern is a dynamic value, you should use Regex.escapeReplacement (see GUIDO's answer).
However, you are replacing a literal value with another literal value, you do not have to use a regex here:
myString.replace("<user_id:int>", """(\d+)""")
See this Kotlin demo yielding /api/(\d+)/.
Note the use of raw string literals where a backslash is parsed as a literal backslash.
The replacement as the regex engine see's it is interpolated as a double quoted string.
This is true with every regex engine.
This is to distinguish control codes, like tab newline or carriage return.
Nothing special here.
So the replacement as the engine wants to see it is (\\d+).
The language interpolates the same.
Final result repl_str = "(\\\\d+)"
I'mm starting to learn about Regular Expressions and I have written code in c++
my task is : Implement a function that replaces each digit in the given string with a '#' character.
For my example, the inputstring = "12 points".
I know I need to use \d for matches a digit. I tried to use this : std::regex_replace(input,std::regex("\d"),"#");
but it is not working: the output is still "12 points";
Then I searched the internet and the result is:
std::regex_replace(input,std::regex("\\d"),"#");
with the output is "## points".
Can anyone help me to understand what is "\\d" ?
\d means decimal, however, in the regular expression, the \ is a special character, which needs to be escaped on its own as well, hence in \\d you escape the \ to mark it to be used as a regular character instead of its special meaning.
When you use "\d" in a C++ application, the \ is an escape character in C++. So it doesn't treat the following d as a d.
Regex then gets a string that doesn't have \d in it, but most likely an empty string (since \d doesn't evaluate to anything in C++ to my knowledge).
When you use "\d" you are escaping the . So C++ reads the string as "\d" as you intended.
An example of when you'd use an escape character, is when you want to output a quote. "\"" would output a single double quote.
I'm working on C++, I'm getting a regular expression from xml file. And i have to search this regular expression in a long string.
e.g. my regular expression is : ".+myFunction"
So if i have to put this regular expression in xml file then i need to use backslach character '\' before '.' in above regular expresssion.
i.e. "\.+myFunction"
Now I'm using boost::regex_search() function to search above regular expression. But due to additional backslash character, function return false.
So how to ignore the backslach character while using boost::regex_search() function??
sample code is as follows:
string longString = "hdh::dfjdj::dfuhgj::myFunction.devide.and"
string regularExp = "\.+myFunction"
const boost::regex searchPattern(regularExp);
if(boost::regex_search(longString, searchPattern))
{
cout <<"Regular expresssion is found" << std::endl;
}
It's not really too clear what you're asking:
In the XML, `".+myFunction" is perfectly legal, as is.
If you're trying to match that exact sequence, you'll need to escape
the backslash twice: once because it has a special meaning in a
string literal, and a second time because it has a special meaning for
regular expressions. You'll also need to escape the quotes, if
they're part of what you're looking for: "\"\\\\.+myFunction\"".
But if you're trying to match an exact sequence, you don't need
regular expressions: std::search is largely sufficient.
If you're trying to define a regular expression which matches a
sequence of one or more characters other than a newline, followed by
the sequence "myFunction", the string literal to initialize the
regular expression would be ".+myFunction".