CWE-1333: Inefficient Regular Expression Complexity
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Edit Custom FilterThe product uses a regular expression with an inefficient, possibly exponential worst-case computational complexity that consumes excessive CPU cycles.
Some regular expression engines have a feature called "backtracking". If the token cannot match, the engine "backtracks" to a position that may result in a different token that can match.
Backtracking becomes a weakness if all of these conditions are met:
Attackers can create crafted inputs that intentionally cause the regular expression to use excessive backtracking in a way that causes the CPU consumption to spike.
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![]() Languages Class: Not Language-Specific (Undetermined Prevalence) Example 1 This example attempts to check if an input string is a "sentence" [REF-1164]. (bad code)
Example Language: JavaScript
var test_string = "Bad characters: $@#";
var bad_pattern = /^(\w+\s?)*$/i; var result = test_string.search(bad_pattern); The regular expression has a vulnerable backtracking clause inside (\w+\s?)*$ which can be triggered to cause a Denial of Service by processing particular phrases. To fix the backtracking problem, backtracking is removed with the ?= portion of the expression which changes it to a lookahead and the \2 which prevents the backtracking. The modified example is: (good code)
Example Language: JavaScript
var test_string = "Bad characters: $@#";
var good_pattern = /^((?=(\w+))\2\s?)*$/i; var result = test_string.search(good_pattern); Note that [REF-1164] has a more thorough (and lengthy) explanation of everything going on within the RegEx. Example 2 This example attempts to check if an input string is a "sentence" and is modified for Perl [REF-1164]. (bad code)
Example Language: Perl
my $test_string = "Bad characters: \$\@\#";
my $bdrslt = $test_string; $bdrslt =~ /^(\w+\s?)*$/i; The regular expression has a vulnerable backtracking clause inside (\w+\s?)*$ which can be triggered to cause a Denial of Service by processing particular phrases. To fix the backtracking problem, backtracking is removed with the ?= portion of the expression which changes it to a lookahead and the \2 which prevents the backtracking. The modified example is: (good code)
Example Language: Perl
my $test_string = "Bad characters: \$\@\#";
my $gdrslt = $test_string; $gdrslt =~ /^((?=(\w+))\2\s?)*$/i; Note that [REF-1164] has a more thorough (and lengthy) explanation of everything going on within the RegEx.
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