Why does the “g” modifier give different results when test() is called twice? [duplicate]

Answer 1

Usually a test is chosen to check if some pattern matches at all, but the global flag lets you loop through a string to either count the matches or,like exec, do something with each lastIndex. Another use is to set the lastIndex of the rx yourself before the test is peformed, to ignore matches before some character index.

var count=0, rx=/\s+/g, rx.lastIndex=100;
while(rx.test(string))count++;

Answer 2

To workaround the problem, you can remove the g flag or reset lastIndex as in

var reg = /a/g;
console.log(reg.test("a"));
reg.lastIndex = 0;
console.log(reg.test("a"));

The problem arises because test is based around exec which looks for more matches after the first if passed the same string and the g flag is present.

15.10.6.3 RegExp.prototype.test(string) # Ⓣ Ⓡ

The following steps are taken:

1. Let match be the result of evaluating the RegExp.prototype.exec (15.10.6.2) algorithm upon this RegExp object using string as the argument.
2. If match is not null, then return true; else return false.

The key part of exec is step 6 of 15.10.6.2:

6. Let global be the result of calling the [[Get]] internal method of R with argument "global".
7. If global is false, then let i = 0.

When i is not reset to 0, then exec (and therefore test) does not start looking at the beginning of the string.

This is useful for exec because you can loop to handle each match:

 var myRegex = /o/g;
 var myString = "fooo";
 for (var match; match = myRegex.exec(myString);) {
   alert(match + " at " + myRegex.lastIndex);
 }

but obviously it isn't so useful for test.