Regular Expressions 101

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An explanation of your regex will be automatically generated as you type.
Detailed match information will be displayed here automatically.
  • All Tokens
  • Common Tokens
  • General Tokens
  • Anchors
  • Meta Sequences
  • Quantifiers
  • Group Constructs
  • Character Classes
  • Flags/Modifiers
  • Substitution
  • A single character of: a, b or c
    [abc]
  • A character except: a, b or c
    [^abc]
  • A character in the range: a-z
    [a-z]
  • A character not in the range: a-z
    [^a-z]
  • A character in the range: a-z or A-Z
    [a-zA-Z]
  • Any single character
    .
  • Alternate - match either a or b
    a|b
  • Any whitespace character
    \s
  • Any non-whitespace character
    \S
  • Any digit
    \d
  • Any non-digit
    \D
  • Any word character
    \w
  • Any non-word character
    \W
  • Non-capturing group
    (?:...)
  • Capturing group
    (...)
  • Zero or one of a
    a?
  • Zero or more of a
    a*
  • One or more of a
    a+
  • Exactly 3 of a
    a{3}
  • 3 or more of a
    a{3,}
  • Between 3 and 6 of a
    a{3,6}
  • Start of string
    ^
  • End of string
    $
  • A word boundary
    \b
  • Non-word boundary
    \B

Regular Expression
No Match

/
/
xgJ

Test String

Code Generator

Language

Generated Code

import java.util.regex.Matcher; import java.util.regex.Pattern; public class Example { public static void main(String[] args) { final String regex = "(?(DEFINE)\n" + "(?<add> \\s*\\+\\s* )\n" + "(?<eq> \\s*=\\s* )\n" + "# Remove all zeroes except the last one if the number is 0\n" + "(?<zero> (?:0(?=\\d))*+ )\n" + "# cl: last digit of left operand being 1, cr: last digit of right operand being 1, \\d(?:0|\\b) check if last digit from result is 0\n" + "# there will be carry if cl and cr are set, or cl or cr are set and the last digit from result is 0\n" + "(?<carry> (?(cl)(?(cr)|\\d(?:0|\\b))|(?(cr)\\d(?:0|\\b)|(*F))) )\n" + "# add carry with l1 (current digit of left operand being 1) and r1 (current digit of right operand being 1)\n" + "# i.e. returns result of carry + l1 + r1 in Z/2Z\n" + "(?<digitadd> (?(?= (?(?=(?(l1)(?(r1)|(*F))|(?(r1)(*F))))(?&carry)|(?!(?&carry))) )1|0) )\n" + "# check for a single digit at the current offset whether the result is correct\n" + "# ro: right operand out of bounds (i.e. the current digit is at a higher offset than the size of the left operand)\n" + "# if we're out of bounds of the right operand, cr is just not set (i.e. handled as if there were leading zeroes)\n" + "(?<recursedigit>\n" + "# now, with the r and f, we can figure out r1 and cr at the current offset and also perform binary carry addition at that offset in the result\n" + " (?&add) (?&zero) (?:\\d*(?:0|1(?<r1>)))? (?(ro)|(?=(?<cr>1)?))\\k<r> (?&eq) \\d*(?&digitadd)\\k<f>\\b\n" + "# iterate through the whole left operand to find the sequences (for right operand and result) of the same length as the offset of the current digit\n" + "| (?=\\d* (?&add) (?&zero) (?:\\k<r>(?<ro>)|\\d*(?<r>\\d\\k<r>)) (?&eq) \\d*(?<f>\\d\\k<f>)\\b) \\d(?&recursedigit)\n" + ")\n" + "# run the check, sets l1 and cl accordingly and initializes the r (right operand) and f (final result) groups to be empty\n" + "(?<checkdigit> (?:0|1(?<l1>)) (?=(?<cl>1)?) (?<r>) (?<f>) (?&recursedigit) )\n" + "# \"trivial\" increment of a binary number, i.e. a +1 is applied to the part of the right operand which exceeds the length of the left operand\n" + "(?<carryoverflow>\n" + "# number contains a zero, just update the part after the last zero\n" + " (?<x>\\d+) 0 (?<y> \\k<r> (?&eq) 0*\\k<x>1 | 1(?&y)0 )\n" + "# number contains only ones, add a leading 1 and replace all the ones by zeroes\n" + "| (?<z> 1\\k<r> (?&eq) 0*10 | 1(?&z)0 )\n" + ")\n" + "# ensure correct lengths of the final operand and handle right operands being longer than the left operand\n" + "(?<recurseoverflow>\n" + "# the left operand is longer than or as long as the right one. In the latter case, the final result will always be exactly one digit longer than the operands\n" + "# in the former case, if the first non-leading zero (from the left) of the left operand is at a higher or equal offset to the length of the right operand, the final result will be one digit longer than the left operand\n" + " (?&add) 0*+ (?(rlast) \\k<r> (?&eq) 0*(?(ro)(?(addone)1)|1)\\k<f>\\b\n" + "# the right operand has a zero at the offset equal to the length of the left operand. Then just copy the leading digits to the final result\n" + " | (?:(?<remaining>\\d+)(?=0\\d* (?&eq) \\d*(?=1)\\k<f>\\b)\\k<r> (?&eq) (*PRUNE) 0*\\k<remaining>\\k<f>\\b\n" + "# otherwise there will be some carry which needs to be applied before copying the leading digits to the final result\n" + " | (?&carryoverflow)\\k<f>\\b))\n" + "# iterate through the whole left operand to find the sequences (for right operand and result) of the same length as the left operand\n" + "| (?=\\d* (?&add) 0*+ (?:\\k<r>(?<ro>)|(?=(?:\\d\\k<r>(?&eq)(?<rlast>))?)\\d*(?<r>\\d\\k<r>)) (?&eq) \\d*(?<f>\\d\\k<f>)\\b)\n" + "# check - only at the first non-leading zero - whether the right operand is longer than the current offset of the iteration, or just as long and having a carry (i.e. the digit at that offset in the final result is 0)\n" + " (?(nullchecked)|(?=(?<addone>(?=0)(?=(?:\\d(?=\\d*(?&add)\\d*(?&eq)\\d*(?<c>\\d\\k<c>)\\b))+(?&add))(?<longer>(?&add)0*|\\d(?&longer)\\d)(\\d+(?&eq)|(?&eq)\\d*(?=0)\\k<c>))?)(?=(?<nullchecked>0)?))\n" + " \\d(?&recurseoverflow)\n" + ")\n" + "(?<s>\n" + " (?=\\d) 0*? (?<arg>[01]+)? (?&add) (?=\\d) 0*? (?<arg>(?(arg)(*F))[01]+)? (?&eq) (*PRUNE) \\k<arg>\n" + "| (?&zero)\n" + "# traverse the digits one by one and verify the correctness of each offset individually\n" + " (?=(?<iteratedigits> (?=(?&checkdigit))\\d (?:\\b|(?&iteratedigits)) ))\n" + "# assert exact format here\n" + " (?=[01]+ (?&add) [01]+ (?&eq) [01]+ \\b)\n" + "# remove leading zeroes and force an additional digit on the final result in case the left operand is only ones and the right operand not longer than the left\n" + " 0*? (?<r>) (?<f>) (?<c>) (?=(?<addone>1+(?&add))?) (?&recurseoverflow)\n" + "# Handle 0 + x or x + 0 separately to avoid messing around in the big subpatterns\n" + ")\n" + ")\n" + "\\b(?&s)\\b"; final String string = "- 1 + 1 = 0\n" + "- 1 + 1 = 1\n" + "- 0 + 1 = 11\n" + "+ 1 + 1 = 10\n" + "- 1 + 10 = 101\n" + "- 0 + 11 = 101\n" + "+ 0 + 1 = 1\n" + "- 0 + 0 = 10\n" + "+ 1 + 0 = 1\n" + "- 1 + 101 = 1010\n" + "+ 10 + 0 = 10\n" + "+ 10 + 1 = 11\n" + "+ 11 + 1 = 100\n" + "+ 100 + 1 = 101\n" + "+ 01 + 10 = 11\n" + "+ 10 + 10 = 100\n" + "+ 010 + 010 = 100\n" + "+ 101 + 11 = 1000\n" + "- 101 + 10 = 1111\n" + "- 100 + 10 = 111\n" + "- 101 + 10 = 101\n" + "+ 101 + 10 = 111\n" + "+ 11 + 101 = 1000\n" + "+ 11 + 1011 = 1110\n" + "+ 11 + 10101 = 11000\n" + "+ 1 + 1111 = 10000\n" + "+ 111 + 11 = 1010\n" + "+ 1110 + 100 = 10010\n" + "+ 11 + 11 = 110\n" + "+ 1111 + 11 = 10010\n" + "- 010 + 010 = 000\n" + "- 100 + 100 = 1100\n" + "+ 100 + 100 = 1000\n" + "+ 1000 + 100 = 1100\n" + "+ 1 + 1000 = 1001\n" + "+ 1000 + 1 = 1001\n" + "+ 110 + 1100 = 10010\n" + "- 110 + 1010 = 1000\n" + "- 10 + 1 = 11111001"; final Pattern pattern = Pattern.compile(regex, Pattern.COMMENTS); final Matcher matcher = pattern.matcher(string); while (matcher.find()) { System.out.println("Full match: " + matcher.group(0)); for (int i = 1; i <= matcher.groupCount(); i++) { System.out.println("Group " + i + ": " + matcher.group(i)); } } } }

Please keep in mind that these code samples are automatically generated and are not guaranteed to work. If you find any syntax errors, feel free to submit a bug report. For a full regex reference for Java, please visit: https://docs.oracle.com/javase/7/docs/api/java/util/regex/Pattern.html