javascript - Why does the "g" modifier give different results when test() is called twice? -

javascript - Why does the "g" modifier give different results when test() is called twice? -

given code:

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

i result:

true false

i have no thought how happen. have tested in both node.js (v8) , firefox browser.

to workaround problem, can remove g flag or reset lastindex in

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

the problem arises because test based around exec looks more matches after first if passed same string , g flag present.

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

the next steps taken:

let match result of evaluating regexp.prototype.exec (15.10.6.2) algorithm upon regexp object using string argument. if match not null, homecoming true; else homecoming false.

the key part of exec step 6 of 15.10.6.2:

6. allow global result of calling [[get]] internal method of r argument "global". 7. if global false, allow = 0.

when i not reset 0, exec (and hence test) not start looking @ origin of string.

this useful exec because can loop handle each match:

var myregex = /o/g; var mystring = "fooo"; (var match; match = myregex.exec(mystring);) { alert(match + " @ " + myregex.lastindex); }

but isn't useful test.

javascript regex