How is \\n and \\\n interpreted by the expanded regular expression? - regex

Within an ERE, a backslash character (\, \a, \b, \f, \n,
\r, \t, \v) is considered to begin an escape sequence.
Then I see \\n and [\\\n], I can guess though both \\n and [\\\n] here means \ followed by new line, but I'm confused by the exact process to interpret such sequence as how many \s are required at all?
UPDATE
I don't have problem understanding regex in programing languages so please make the context within the lexer.
[root# ]# echo "test\
> hi"

This is dependent on the programming language and on its string handling options.
For example, in Java strings, if you need a literal backslash in a string, you need to double it. So the regex \n must be written as "\\n". If you plan to match a backslash using a regex, then you need to escape it twice - once for Java's string handler, and once for the regex engine. So, to match \, the regex is \\, and the corresponding Java string is "\\\\".
Many programming languages have special "verbatim" or "raw" strings where you don't need to escape backslashes. So the regex \n can be written as a normal Python string as "\\n" or as a Python raw string as r"\n". The Python string "\n" is the actual newline character.
This can becoming confusing, because sometimes not escaping the backslash happens to work. For example the Python string "\d\n" happens to work as a regex that's intended to match a digit, followed by a newline. This is because \d isn't a recognized character escape sequence in Python strings, so it's kept as a literal \d and fed that way to the regex engine. The \n is translated to an actual newline, but that happens to match the newline in the string that the regex is tested against.
However, if you forget to escape a backslash where the resulting sequence is a valid character escape sequence, bad things happen. For example, the regex \bfoo\b matches an entire word foo (but it doesn't match the foo in foobar). If you write the regex string as "\bfoo\b", the \bs are translated into backspace characters by the string processor, so the regex engine is told to match <backspace>foo<backspace> which obviously will fail.
Solution: Always use verbatim strings where you have them (e. g. Python's r"...", .NET's #"...") or use regex literals where you have them (e. g. JavaScript's and Ruby's /.../). Or use RegexBuddy to automatically translate the regex for you into your language's special format.
To get back to your examples:
\\n as a regex means "Match a backslash, followed by n"
[\\\n] as a regex means "Match either a backslash or a newline character".

Actually regex string specified by string literal is processed by two compilers: programming language compiler and regexp compiler:
Original Compiled Regex compiled
"\n" NL NL
"\\n" '\'+'n' NL
"\\\n" '\'+NL NL
"\\\\n" '\'+'\'+'n' '\'+'n'
So you must use the shortest format "\n".
Code examples:
JavaScript:
'a\nb'.replace(RegExp("\n"),'<br>')
'a\nb'.replace(RegExp("\\n"),'<br>')
'a\nb'.replace(RegExp("\\\n"),'<br>')
but not:
'a\nb'.replace(/\\\n/,'<br>')
Java:
System.out.println("a\nb".replaceAll("\n","<br>"));
System.out.println("a\nb".replaceAll("\\n","<br>"));
System.out.println("a\nb".replaceAll("\\\n","<br>"));
Python:
str.join('<br>',regex.split('\n','a\nb'))
str.join('<br>',regex.split('\\n','a\nb'))
str.join('<br>',regex.split('\\\n','a\nb'))

Related

How to exclude part of string using regex and change add this part and the and of string?

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.)

Why do I need to write \\d instead of \d in a C++ regex?

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.

Regex: A metacharacter to recognize any type of character

I want to be able to recognize the following pattern:
$...$
with "..." being any type of character.
How can it be done in Kotlin?
The dollar sign ($) is a boundary matcher for the end of a line so you have to escape it with backslash (\). Kotlin has two types of strings: escaped and raw (see Basic Types - Kotlin Programming Language). If you use escaped strings then you'll need to escape the backslash with a backslash. Below are various ways to declare your desired Regex:
"""\$.{3}\$""".toRegex()
"""\$...\$""".toRegex()
"\\$.{3}\\$".toRegex()
"\\$...\\$".toRegex()
Regex("""\$.{3}\$""")
Regex("""\$...\$""")
Regex("\\$.{3}\\$")
Regex("\\$...\\$")

In postgreSQL, why is \s treated differently from \w?

Here is the example that confuses me:
select ' w' ~ '^\s\w$';
This results in "false", but seems like it should be true.
select ' w' ~ '^\\s\w*$';
This results in "true", but:
Why does \s need the extra backslash?
If it truly does, why does \w not need the extra backslash?
Thanks for any help!
I think you have tested it the wrong way because I'm getting the opposite results that you got.
select ' w' ~ '^\s\w$';
Is returning 1 in my case. Which actually makes sense because it is matching the space at the beginning of the text, followed by the letter at the end.
select ' w' ~ '^\\s\w*$';
Is returning 0 and it makes sense too. Here you're trying to match a backslash at the beginning of the text followed by an s and then, by any number of letters, numbers or underscores.
A piece of text that would match your second regex would be: '\sw'
Check the fiddle here.
The string constants are first parsed and interpreted as strings, including escaped characters. Escaping of unrecognized sequences is handled differently by different parsers, but generally, besides errors, the most common behavior is to ignore the backslash.
In the first example, the right-hand string constant is first being interpreted as '^sw$', where both \s and \w are not recognized string escape sequences.
In the second example the right hand constant is interpreted as '^\sw*$' where \\s escapes the \
After the strings are interpreted they are then applied as a regular expression, '^\sw*$' matching ' w' where '^sw$' does not.
Some languages use backslash as an escape character. Regexes do that, C-like languages do that, and some rare and odd dialects of SQL do that. PostgresSQL does it. PostgresSQL is translating the backslash escaping to arrive at a string value, and then feeding that string value to the regex parser, which AGAIN translates whatever backslashes survived the first translation -- if any. In your first regex, none did.
For example, in a string literal or a regex, \n doesn't mean a backslash followed by a lowercase n. It means a newline. Depending on the language, a backslash followed by a lowercase s will mean either just a lowercase s, or nothing. In PostgresSQL, an invalid escape sequence in a string literal translates as the escaped character: '\w' translates to 'w'. All the regex parser sees there is the w. By chance, you used the letter w in the string you're matching against. It's not matching that w in the lvalue because it's a word character; it's matching it because it's a lowercase w. Change it to lowercase x and it'll stop matching.
If you want to put a backslash in a string literal, you need to escape it with another backslash: '\\'. This is why \\s in your second regex worked. Add a second backslash to \w if you want to match any word character with that one.
This is a horrible pain. It's why JavaScript, Perl, and other languages have special conventions for regex literals like /\s\w/, and why C# programmers use the #"string literal" feature to disable backslash escaping in strings they intend to use as regexes.

c++ regexp for not preceded by backslash and preceded by backslash

I can only find negative lookbehind for this , something like (?<!\\).
But this won't compile in c++ and flex. It seems like both regex.h nor flex support this?
I am trying to implement a shell which has to get treat special char like >, < of | as normal argument string if preceded by backslash. In other word, only treat special char as special if not preceded by 0 or even number of '\'
So echo \\>a or echo abc>a should direct output to a
but echo \>a should print >a
What regular expression should I use?
I'm using flex and yacc to parse the input.
In a Flex rule file, you'd use \\ to match a single backslash '\' character. This is because the \ is used as an escape character in Flex.
BACKSLASH \\
LITERAL_BACKSLASH \\\\
LITERAL_LESSTHAN \\\\<
LITERAL_GREATERTHAN \\\\>
LITERAL_VERTICALBAR \\\\|
If I follow you correctly, in your case you want "\>" to be treated as literal '>' but "\\>" to be treated as literal '\' followed by special redirect. You don't need negative look behind or anything particularly special to accomplish this as you can build one rule that would accept both your regular argument characters and also the literal versions of your special characters.
For purposes of discussion, let's assume that your argument/parameter can contain any character but ' ', '\t', and the special forms of '>', '<', '|'. The rule for the argument would then be something like:
ARGUMENT ([^ \t\\><|]|\\\\|\\>|\\<|\\\|)+
Where:
[^ \t\\><|] matches any single character but ' ', '\t', and your special characters
\\\\ matches any instance of "\" (i.e. a literal backslash)
\\> matches any instance of ">" (i.e. a literal greater than)
\\< matches any instance of "\<" (i.e. a literal less than)
\\\| matches any instance of "\|" (i.e. a literal vertical bar/pipe)
Actually... You can probably just shorten that rule to:
ARGUMENT ([^ \t\\><|]|\\[^ \t\r\n])+
Where:
[^ \t\\><|] matches any single character but ' ', '\t', and your special characters
\\[^ \t\r\n] matches any character preceded by a '\' in your input except for whitespace (which will handle all of your special characters and allow for literal forms of all other characters)
If you want to allow for literal whitespace in your arguments/parameters then you could shorten the rule even further but be careful with using \\. for the second half of the rule alternation as it may or may not match " \n" (i.e. eat your trailing command terminator character!).
Hope that helps!
You cannot easily extract single escaped characters from a command-line, since you will not know the context of the character. In the simplest case, consider the following:
LessThan:\<
BackslashFrom:\\<
In the first one, < is an escaped character; in the second one, it is not. If your language includes quotes (as most shells do), things become even more complicated. It's a lot better to parse the string left to right, one entity at a time. (I'd use flex myself, because I've stopped wasting my time writing and testing lexers, but you might have some pedagogical reason to do so.)
If you really need to find a special character which shouldn't be special, just search for it (in C++98, where you don't have raw literals, you'll have to escape all of the backslashes):
regex: (\\\\)*\\[<>|]
(An even number -- possibly 0 -- of \, then a \ and a <, > or |)
as a C string => "(\\\\\\\\)*\\\\[<>|]"