Reputation: 79826
What are these two terms in an understandable way?
Upvotes: 843
Views: 509764
Reputation: 35731
They’ll take as much as they can. e.g. matches with this regex: .*
$50,000
Kiss your bank balance goodbye with that regex!
See here for an example: Greedy-example
Ask for a tax refund: the IRS sudden becomes non-greedy - and return as little as possible: i.e. they use this quantifier:
(.{2,5}?)([0-9]*)
against this input: $50,000
The first group is non-needy and only matches $5
– so I get a $5
refund against the $50,000 input.
See here: Non-greedy-example.
It becomes important if you are trying to match certain parts of an expression. Sometimes you don't want to match everything - as little as possible. Sometimes you want to match as much as possible. Nothing more to it.
You can play around with the examples in the links posted above.
(Analogies used to help you remember.)
Upvotes: 9
Reputation: 268462
Greedy will consume as much as possible. From http://www.regular-expressions.info/repeat.html we see the example of trying to match HTML tags with <.+>
. Suppose you have the following:
<em>Hello World</em>
You may think that <.+>
(.
means any non newline character and +
means one or more) would only match the <em>
and the </em>
, when in reality it will be very greedy, and go from the first <
to the last >
. This means it will match <em>Hello World</em>
instead of what you wanted.
Making it lazy (<.+?>
) will prevent this. By adding the ?
after the +
, we tell it to repeat as few times as possible, so the first >
it comes across, is where we want to stop the matching.
By the way, I suggest you download RegExr, a great tool that will help you explore regular expressions - I use it all the time.
Upvotes: 981
Reputation: 11085
As far as I know, most regex engine is greedy by default. Add a question mark at the end of quantifier will enable lazy match.
As @Andre S mentioned in comment.
Refer to the example below for what is greedy and what is lazy.
import java.util.regex.Matcher;
import java.util.regex.Pattern;
public class Test {
public static void main(String args[]){
String money = "100000000999";
String greedyRegex = "100(0*)";
Pattern pattern = Pattern.compile(greedyRegex);
Matcher matcher = pattern.matcher(money);
while(matcher.find()){
System.out.println("I'm greedy and I want " + matcher.group() + " dollars. This is the most I can get.");
}
String lazyRegex = "100(0*?)";
pattern = Pattern.compile(lazyRegex);
matcher = pattern.matcher(money);
while(matcher.find()){
System.out.println("I'm too lazy to get so much money, only " + matcher.group() + " dollars is enough for me");
}
}
}
I'm greedy and I want 100000000 dollars. This is the most I can get.
I'm too lazy to get so much money, only 100 dollars is enough for me
Upvotes: 29
Reputation: 449
To give extra clarification on Laziness, here is one example which is maybe not intuitive on first look but explains idea of "gradually expands the match" from Suganthan Madhavan Pillai answer.
input -> [email protected]@
regex -> ^.*?@$
Regex for this input will have a match. At first glance somebody could say LAZY match(".*?@") will stop at first @ after which it will check that input string ends("$"). Following this logic someone would conclude there is no match because input string doesn't end after first @.
But as you can see this is not the case, regex will go forward even though we are using non-greedy(lazy mode) search until it hits second @ and have a MINIMAL match.
Upvotes: 0
Reputation: 74671
Greedy quantifier | Lazy quantifier | Description |
---|---|---|
* |
*? |
Star Quantifier: 0 or more |
+ |
+? |
Plus Quantifier: 1 or more |
? |
?? |
Optional Quantifier: 0 or 1 |
{n} |
{n}? |
Quantifier: exactly n |
{n,} |
{n,}? |
Quantifier: n or more |
{n,m} |
{n,m}? |
Quantifier: between n and m |
Add a ? to a quantifier to make it ungreedy i.e lazy.
Example:
test string : stackoverflow
greedy reg expression : s.*o
output: stackoverflow
lazy reg expression : s.*?o
output: stackoverflow
Upvotes: 305
Reputation: 69
Best shown by example. String. 192.168.1.1
and a greedy regex \b.+\b
You might think this would give you the 1st octet but is actually matches against the whole string. Why? Because the.+ is greedy and a greedy match matches every character in 192.168.1.1
until it reaches the end of the string. This is the important bit! Now it starts to backtrack one character at a time until it finds a match for the 3rd token (\b
).
If the string a 4GB text file and 192.168.1.1 was at the start you could easily see how this backtracking would cause an issue.
To make a regex non greedy (lazy) put a question mark after your greedy search e.g
*?
??
+?
What happens now is token 2 (+?
) finds a match, regex moves along a character and then tries the next token (\b
) rather than token 2 (+?
). So it creeps along gingerly.
Upvotes: 6
Reputation: 1608
Greedy means it will consume your pattern until there are none of them left and it can look no further.
Lazy will stop as soon as it will encounter the first pattern you requested.
One common example that I often encounter is \s*-\s*?
of a regex ([0-9]{2}\s*-\s*?[0-9]{7})
The first \s*
is classified as greedy because of *
and will look as many white spaces as possible after the digits are encountered and then look for a dash character "-". Where as the second \s*?
is lazy because of the present of *?
which means that it will look the first white space character and stop right there.
Upvotes: 5
Reputation: 2113
Greedy matching. The default behavior of regular expressions is to be greedy. That means it tries to extract as much as possible until it conforms to a pattern even when a smaller part would have been syntactically sufficient.
Example:
import re
text = "<body>Regex Greedy Matching Example </body>"
re.findall('<.*>', text)
#> ['<body>Regex Greedy Matching Example </body>']
Instead of matching till the first occurrence of ‘>’, it extracted the whole string. This is the default greedy or ‘take it all’ behavior of regex.
Lazy matching, on the other hand, ‘takes as little as possible’. This can be effected by adding a ?
at the end of the pattern.
Example:
re.findall('<.*?>', text)
#> ['<body>', '</body>']
If you want only the first match to be retrieved, use the search method instead.
re.search('<.*?>', text).group()
#> '<body>'
Source: Python Regex Examples
Upvotes: 10
Reputation: 114014
'Greedy' means match longest possible string.
'Lazy' means match shortest possible string.
For example, the greedy h.+l
matches 'hell'
in 'hello'
but the lazy h.+?l
matches 'hel'
.
Upvotes: 476
Reputation: 1773
try to understand the following behavior:
var input = "0014.2";
Regex r1 = new Regex("\\d+.{0,1}\\d+");
Regex r2 = new Regex("\\d*.{0,1}\\d*");
Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // "0014.2"
input = " 0014.2";
Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // " 0014"
input = " 0014.2";
Console.WriteLine(r1.Match(input).Value); // "0014.2"
Console.WriteLine(r2.Match(input).Value); // ""
Upvotes: -3
Reputation: 5655
Taken From www.regular-expressions.info
Greediness: Greedy quantifiers first tries to repeat the token as many times as possible, and gradually gives up matches as the engine backtracks to find an overall match.
Laziness: Lazy quantifier first repeats the token as few times as required, and gradually expands the match as the engine backtracks through the regex to find an overall match.
Upvotes: 25
Reputation: 225232
Greedy means your expression will match as large a group as possible, lazy means it will match the smallest group possible. For this string:
abcdefghijklmc
and this expression:
a.*c
A greedy match will match the whole string, and a lazy match will match just the first abc
.
Upvotes: 79
Reputation: 166606
From Regular expression
The standard quantifiers in regular expressions are greedy, meaning they match as much as they can, only giving back as necessary to match the remainder of the regex.
By using a lazy quantifier, the expression tries the minimal match first.
Upvotes: 10