I am having a hard time with this problem.
"Write a flex code which recognizes a chain with alphabet {0,1}, with at least 5 char's, and to every consecutive 5 char's there will bee at least 3 1's"
I thought I have solved, but I am new using flex, so I am getting this "flex scanner push-back overflow".
here's my code
%{
#define ACCEPT 1
#define DONT 2
%}
delim [ \t\n\r]
ws {delim}+
comb01 00111|{comb06}1
comb02 01011|{comb07}1
comb03 01101|{comb08}1
comb04 01110|({comb01}|{comb09})0
comb05 01111|({comb01}|{comb09})1
comb06 10011|{comb10}1
comb07 10101|{comb11}1
comb08 10110|({comb02}|{comb12})0
comb09 10111|({comb02}|{comb12})1
comb10 11001|{comb13}1
comb11 11010|({comb03}|{comb14})0
comb12 11011|({comb03}|{comb14})1
comb13 11100|({comb04}|{comb15})0
comb14 11101|({comb04}|{comb15})1
comb15 11110|({comb05}|{comb16})0
comb16 11111|({comb05}|{comb16})1
accept {comb01}|{comb02}|{comb03}|{comb04}|{comb05}|{comb06}|{comb07}|{comb08}|{comb09}|{comb10}|{comb11}|{comb12}|{comb13}|{comb14}|{comb15}|{comb16}
string [^ \t\n\r]+
%%
{ws} { ;}
{accept} {return ACCEPT;}
{string} {return DONT;}
%%
void main () {
int i;
while (i = yylex ())
switch (i) {
case ACCEPT:
printf ("%-20s: ACCEPT\n", yytext);
break;
case DONT:
printf ("%-20s: Reject\n", yytext);
break;
}
}
Flex definitions are macros, and flex implements them that way: when it sees {defn} in a pattern, it replaces it with whatever defn was defined as (in parentheses, usually, to avoid operator precedence issues). It doesn't expand the macros in the macro definition, so the macro substitution might contain more definition references which in turn need to be substituted.
Since macro substitution is unconditional, it is not possible to use recursive macros, including macros which are indirectly recursive. Which yours are. Flex doesn't check for this condition, unlike the C preprocessor; it just continues substituting in an endless loop until it runs out of space.
(Flex is implemented using itself; it does the macro substitution using unput. unput will not resize the input buffer, so "runs out of space" here means that flex's internal flex's input buffer became full of macro substitutions.)
The strategy you are using would work fine as a context-free grammar. But that's not flex. Flex is about regular expressions. The pattern you want to match can be described by a regular expression -- the "grammar" you wrote with flex macros is a regular grammar -- but it is not a regular expression and flex won't make one out of it for you, unfortunately. That's your job.
I don't think it's going to be a very pretty regular expression. In fact, I think it's likely to be enormous. But I didn't try working it out..
There are flex tricks you could use to avoid constructing the regular expression. For example, you could build your state machine out of flex start conditions and then scan one character at a time, where each character scanned does a state transition or throws an error. (Use more() if you want to return the entire string scanned at the end.)
https://regex101.com/r/sB9wW6/1
(?:(?<=\s)|^)#(\S+) <-- the problem in positive lookbehind
Working like this on prod: (?:\s|^)#(\S+), but I need a correct start index (without space).
Here is in JS:
var regex = new RegExp(/(?:(?<=\s)|^)#(\S+)/g);
Error parsing regular expression: Invalid regular expression:
/(?:(?<=\s)|^)#(\S+)/
What am I doing wrong?
UPDATE
Ok, no lookbehind in JS :(
But anyways, I need a regex to get the proper start and end index of my match. Without leading space.
Make sure you always select the right regex engine at regex101.com. See an issue that occurred due to using a JS-only compatible regex with [^] construct in Python.
JS regex - at the time of answering this question - did not support lookbehinds. Now, it becomes more and more adopted after its introduction in ECMAScript 2018. You do not really need it here since you can use capturing groups:
var re = /(?:\s|^)#(\S+)/g;
var str = 's #vln1\n#vln2\n';
var res = [];
while ((m = re.exec(str)) !== null) {
res.push(m[1]);
}
console.log(res);
The (?:\s|^)#(\S+) matches a whitespace or the start of string with (?:\s|^), then matches #, and then matches and captures into Group 1 one or more non-whitespace chars with (\S+).
To get the start/end indices, use
var re = /(\s|^)#\S+/g;
var str = 's #vln1\n#vln2\n';
var pos = [];
while ((m = re.exec(str)) !== null) {
pos.push([m.index+m[1].length, m.index+m[0].length]);
}
console.log(pos);
BONUS
My regex works at regex101.com, but not in...
First of all, have you checked the Code Generator link in the Tools pane on the left?
All languages - "Literal string" vs. "String literal" alert - Make sure you test against the same text used in code, literal string, at the regex tester. A common scenario is copy/pasting a string literal value directly into the test string field, with all string escape sequences like \n (line feed char), \r (carriage return), \t (tab char). See Regex_search c++, for example. Mind that they must be replaced with their literal counterparts. So, if you have in Python text = "Text\n\n abc", you must use Text, two line breaks, abc in the regex tester text field. Text.*?abc will never match it although you might think it "works". Yes, . does not always match line break chars, see How do I match any character across multiple lines in a regular expression?
All languages - Backslash alert - Make sure you correctly use a backslash in your string literal, in most languages, in regular string literals, use double backslash, i.e. \d used at regex101.com must written as \\d. In raw string literals, use a single backslash, same as at regex101. Escaping word boundary is very important, since, in many languages (C#, Python, Java, JavaScript, Ruby, etc.), "\b" is used to define a BACKSPACE char, i.e. it is a valid string escape sequence. PHP does not support \b string escape sequence, so "/\b/" = '/\b/' there.
All languages - Default flags - Global and Multiline - Note that by default m and g flags are enabled at regex101.com. So, if you use ^ and $, they will match at the start and end of lines correspondingly. If you need the same behavior in your code check how multiline mode is implemented and either use a specific flag, or - if supported - use an inline (?m) embedded (inline) modifier. The g flag enables multiple occurrence matching, it is often implemented using specific functions/methods. Check your language reference to find the appropriate one.
line-breaks - Line endings at regex101.com are LF only, you can't test strings with CRLF endings, see regex101.com VS myserver - different results. Solutions can be different for each regex library: either use \R (PCRE, Java, Ruby) or some kind of \v (Boost, PCRE), \r?\n, (?:\r\n?|\n)/(?>\r\n?|\n) (good for .NET) or [\r\n]+ in other libraries (see answers for C#, PHP). Another issue related to the fact that you test your regex against a multiline string (not a list of standalone strings/lines) is that your patterns may consume the end of line, \n, char with negated character classes, see an issue like that. \D matched the end of line char, and in order to avoid it, [^\d\n] could be used, or other alternatives.
php - You are dealing with Unicode strings, or want shorthand character classes to match Unicode characters, too (e.g. \w+ to match Стрибижев or Stribiżew, or \s+ to match hard spaces), then you need to use u modifier, see preg_match() returns 0 although regex testers work - To match all occurrences, use preg_match_all, not preg_match with /...pattern.../g, see PHP preg_match to find multiple occurrences and "Unknown modifier 'g' in..." when using preg_match in PHP?- Your regex with inline backreference like \1 refuses to work? Are you using a double quoted string literal? Use a single-quoted one, see Backreference does not work in PHP
phplaravel - Mind you need the regex delimiters around the pattern, see https://stackoverflow.com/questions/22430529
python - Note that re.search, re.match, re.fullmatch, re.findall and re.finditer accept the regex as the first argument, and the input string as the second argument. Not re.findall("test 200 300", r"\d+"), but re.findall(r"\d+", "test 200 300"). If you test at regex101.com, please check the "Code Generator" page. - You used re.match that only searches for a match at the start of the string, use re.search: Regex works fine on Pythex, but not in Python - If the regex contains capturing group(s), re.findall returns a list of captures/capture tuples. Either use non-capturing groups, or re.finditer, or remove redundant capturing groups, see re.findall behaves weird - If you used ^ in the pattern to denote start of a line, not start of the whole string, or used $ to denote the end of a line and not a string, pass re.M or re.MULTILINE flag to re method, see Using ^ to match beginning of line in Python regex
- If you try to match some text across multiple lines, and use re.DOTALL or re.S, or [\s\S]* / [\s\S]*?, and still nothing works, check if you read the file line by line, say, with for line in file:. You must pass the whole file contents as the input to the regex method, see Getting Everything Between Two Characters Across New Lines. - Having trouble adding flags to regex and trying something like pattern = r"/abc/gi"? See How to add modifers to regex in python?
c#, .net - .NET regex does not support possessive quantifiers like ++, *+, ??, {1,10}?, see .NET regex matching digits between optional text with possessive quantifer is not working - When you match against a multiline string and use RegexOptions.Multiline option (or inline (?m) modifier) with an $ anchor in the pattern to match entire lines, and get no match in code, you need to add \r? before $, see .Net regex matching $ with the end of the string and not of line, even with multiline enabled - To get multiple matches, use Regex.Matches, not Regex.Match, see RegEx Match multiple times in string - Similar case as above: splitting a string into paragraphs, by a double line break sequence - C# / Regex Pattern works in online testing, but not at runtime - You should remove regex delimiters, i.e. #"/\d+/" must actually look like #"\d+", see Simple and tested online regex containing regex delimiters does not work in C# code - If you unnecessarily used Regex.Escape to escape all characters in a regular expression (like Regex.Escape(#"\d+\.\d+")) you need to remove Regex.Escape, see Regular Expression working in regex tester, but not in c#
dartflutter - Use raw string literal, RegExp(r"\d"), or double backslashes (RegExp("\\d")) - https://stackoverflow.com/questions/59085824
javascript - Double escape backslashes in a RegExp("\\d"): Why do regex constructors need to be double escaped?
- (Negative) lookbehinds unsupported by most browsers: Regex works on browser but not in Node.js - Strings are immutable, assign the .replace result to a var - The .replace() method does change the string in place - Retrieve all matches with str.match(/pat/g) - Regex101 and Js regex search showing different results or, with RegExp#exec, RegEx to extract all matches from string using RegExp.exec- Replace all pattern matches in string: Why does javascript replace only first instance when using replace?
javascriptangular - Double the backslashes if you define a regex with a string literal, or just use a regex literal notation, see https://stackoverflow.com/questions/56097782
java - Word boundary not working? Make sure you use double backslashes, "\\b", see Regex \b word boundary not works - Getting invalid escape sequence exception? Same thing, double backslashes - Java doesn't work with regex \s, says: invalid escape sequence - No match found is bugging you? Run Matcher.find() / Matcher.matches() - Why does my regex work on RegexPlanet and regex101 but not in my code? - .matches() requires a full string match, use .find(): Java Regex pattern that matches in any online tester but doesn't in Eclipse - Access groups using matcher.group(x): Regex not working in Java while working otherwise - Inside a character class, both [ and ] must be escaped - Using square brackets inside character class in Java regex - You should not run matcher.matches() and matcher.find() consecutively, use only if (matcher.matches()) {...} to check if the pattern matches the whole string and then act accordingly, or use if (matcher.find()) to check if there is a single match or while (matcher.find()) to find multiple matches (or Matcher#results()). See Why does my regex work on RegexPlanet and regex101 but not in my code?
scala - Your regex attempts to match several lines, but you read the file line by line (e.g. use for (line <- fSource.getLines))? Read it into a single variable (see matching new line in Scala regex, when reading from file)
kotlin - You have Regex("/^\\d+$/")? Remove the outer slashes, they are regex delimiter chars that are not part of a pattern. See Find one or more word in string using Regex in Kotlin - You expect a partial string match, but .matchEntire requires a full string match? Use .find, see Regex doesn't match in Kotlin
mongodb - Do not enclose /.../ with single/double quotation marks, see mongodb regex doesn't work
c++ - regex_match requires a full string match, use regex_search to find a partial match - Regex not working as expected with C++ regex_match - regex_search finds the first match only. Use sregex_token_iterator or sregex_iterator to get all matches: see What does std::match_results::size return? - When you read a user-defined string using std::string input; std::cin >> input;, note that cin will only get to the first whitespace, to read the whole line properly, use std::getline(std::cin, input); - C++ Regex to match '+' quantifier - "\d" does not work, you need to use "\\d" or R"(\d)" (a raw string literal) - This regex doesn't work in c++ - Make sure the regex is tested against a literal text, not a string literal, see Regex_search c++
go - Double backslashes or use a raw string literal: Regular expression doesn't work in Go - Go regex does not support lookarounds, select the right option (Go) at regex101.com before testing! Regex expression negated set not working golang
groovy - Return all matches: Regex that works on regex101 does not work in Groovy
r - Double escape backslashes in the string literal: "'\w' is an unrecognized escape" in grep - Use perl=TRUE to PCRE engine ((g)sub/(g)regexpr): Why is this regex using lookbehinds invalid in R?
oracle - Greediness of all quantifiers is set by the first quantifier in the regex, see Regex101 vs Oracle Regex (then, you need to make all the quantifiers as greedy as the first one)] - \b does not work? Oracle regex does not support word boundaries at all, use workarounds as shown in Regex matching works on regex tester but not in oracle
firebase - Double escape backslashes, make sure ^ only appears at the start of the pattern and $ is located only at the end (if any), and note you cannot use more than 9 inline backreferences: Firebase Rules Regex Birthday
firebasegoogle-cloud-firestore - In Firestore security rules, the regular expression needs to be passed as a string, which also means it shouldn't be wrapped in / symbols, i.e. use allow create: if docId.matches("^\\d+$").... See https://stackoverflow.com/questions/63243300
google-data-studio - /pattern/g in REGEXP_REPLACE must contain no / regex delimiters and flags (like g) - see How to use Regex to replace square brackets from date field in Google Data Studio?
google-sheets - If you think REGEXEXTRACT does not return full matches, truncates the results, you should check if you have redundant capturing groups in your regex and remove them, or convert the capturing groups to non-capturing by add ?: after the opening (, see Extract url domain root in Google Sheet
sed - Why does my regular expression work in X but not in Y?
word-boundarypcrephp - [[:<:]] and [[:>:]] do not work in the regex tester, although they are valid constructs in PCRE, see https://stackoverflow.com/questions/48670105
snowflake-cloud-data-platform snowflake-sql - If you are writing a stored procedure, and \\d does not work, you need to double them again and use \\\\d, see REGEX conversion of VARCHAR value to DATE in Snowflake stored procedure using RLIKE not consistent.
I am using Clang/libtooling (ASTComsumer with a Matcher) to visit ALL return statements (ReturnStmt). I need to extract the expression that comes after the keyword return in a string form so that I can put that in a macro that I am replacing return statement with.
For example, I want to replace the following line:
return somefunc() + 1;
with
FUNCTION_EXIT(somefunc() + 1); // FUNCTION_EXIT is a C macro
The macro will return from the function after doing some logging.
I am using ReturnStmt::getRetValue() that returns an Expr and tried to get it in string form (so that it can be passed to the macro), but I haven't found a way yet. Is there a way to stringify Expr?
Clang has a strict separation of concerns between the abstract syntax tree (AST) and the actual source code. The component that converts between these is the Lexer. To get the raw source for an Expr e:
const string text = Lexer::getSourceText(e.getSourceRange(), source_manager, opt);
Note that the SourceManager and LangOptions are available from the ASTContext. If the code you're parsing has macros then things get more complicated because you have to care about spelling location versus expansion location; SourceManager has a bunch of different functions to convert between these.
Good luck!
Is there a way to use a heredoc type of notation in Lua that references variables within the string?
The basic idea of what I'm trying to do would look something like follows. The heredoc piece is fine, but within Lua you can't actually reference the variable in the manner I'm showing below.
username = "bigtunacan"
sql=[=[
SELECT * FROM users WHERE username='$bigtunacan';
]=]
There's no built-in string interpolation, but it can be trivially implemented with gsub and replacement table.
sql=[=[
SELECT * FROM users WHERE username='$username';
]=]
print((sql:gsub('$(%w+)', { username = 'bigtucan' })))
-- SELECT * FROM users WHERE username='bigtucan';
Note an extra set of () - this is so only first return - the interpolated string is used from gsub and the 2nd - number of replacements made - silently discarded. This might be important if you use result of gsub as last in list of arguments to some function where adding one more argument might produce different behavior.
Also if you want to use this in SQL context, you really should use placeholders instead.
There is no Lua construct that allows variable interpolation within any string. See Literal Strings in the official reference guide.
You could of course write a function that would parse it and do the substitutions.
I noticed that the following code compiles and works in VS 2013:
let f() =
do Console.WriteLine(41)
42
But when looking at the F# 3.0 specification I can't find any mention of do being used this way. As far as I can tell, do can have the following uses:
As a part of loop (e.g. while expr do expr done), that's not the case here.
Inside computation expressions, e.g.:
seq {
for i in 1..2 do
do Console.WriteLine(i)
yield i * 2
}
That's not the case here either, f doesn't contain any computation expressions.
Though what confuses me here is that according to the specification, do should be followed by in. That in should be optional due to lightweight syntax, but adding it here causes a compile error (“Unexpected token 'in' or incomplete expression”).
Statement inside a module or class. This is also not the case here, the do is inside a function, not inside a module or a class.
I also noticed that with #light "off", the code doesn't compile (“Unexpected keyword 'do' in binding”), but I didn't find anything that would explain this in the section on lightweight syntax either.
Based on all this, I would assume that using do inside a function this way should not compile, but it does. Did I miss something in the specification? Or is this actually a bug in the compiler or in the specification?
From the documentation on MSDN:
A do binding is used to execute code without defining a function or value.
Even though the spec doesn't contain a comprehensive list of the places it is allowed, it is merely an expression asserted to be of type unit. Some examples:
if ((do ()); true) then ()
let x: unit = do ()
It is generally omitted. Each of the preceding examples are valid without do. Therefore, do serves only to assert that an expression is of type unit.
Going through the F# 3.0 specification expression syntax has do expr as a choice of class-function-or-value-defn (types) [Ch 8, A.2.5] and module-function-or-value-defn (modules) [Ch 10, A.2.1.1].
I don't actually see in the spec where function-defn can have more than one expression, as long all but the last one evaluate to unit -- or that all but the last expression is ignored in determining the functions return value.
So, it seems this is an oversight in the documentation.