Question

I have made a simple code for capturing a certain group in a string :

/[a-z]+([0-9]+)[a-z]+/gi    (n chars , m digts , k chars).

code :

var myString='aaa111bbb222ccc333ddd';
var myRegexp=/[a-z]+([0-9]+)[a-z]+/gi;

var match=myRegexp.exec(myString);
console.log(match)
 
 while (match != null)
{
  match = myRegexp.exec(myString);
  console.log(match)
}

The result were :

["aaa111bbb", "111"]
["ccc333ddd", "333"]
null

But wait a minute ,
Why he didnt try the bbb222ccc part ?

I mean ,
It saw the aaa111bbb but then he should have try the bbb222ccc… ( That’s greedy !)

What am I missing ?

Also

looking at

   while (match != null)
    {
      match = myRegexp.exec(myString);
      console.log(match)
    }

how did it progressed to the second result ?
at first there was :

var match=myRegexp.exec(myString);

later ( in a while loop)

match=myRegexp.exec(myString);
match=myRegexp.exec(myString);

it is the same line … where does it remember that the first result was already shown ?

Answer 1

.exec is stateful when you use the g flag. The state is kept in the regex object’s .lastIndex property.

var myString = 'aaa111bbb222ccc333ddd';
var myRegexp = /[a-z]+([0-9]+)[a-z]+/gi;
var match = myRegexp.exec(myString);
console.log(myRegexp.lastIndex); //9, so the next `.exec` will only look after index 9
while (match != null) {
    match = myRegexp.exec(myString);
    console.log(myRegexp.lastIndex);
}

The state can be resetted by setting .lastIndex to 0 or by execing a different string. re.exec("") for instance will reset the state because the state was kept for 'aaa111bbb222ccc333ddd'.

The same applies to .test method as well, so never use g flag with a regex that is used for .test if you prefer no surprises. See https://developer.mozilla.org/en-US/docs/JavaScript/Reference/Global_Objects/RegExp/exec