为什么在JavaScript中反转全局正则表达式总是返回true

Question

Sample Code

This works as advertised:

var re = /abc/;
re.test("abc");
//true
!re.test("abc");
//false;

This does not:

var re2 = /abc/g;
//undefined
re2.test("abc");
//true
!re2.test("abc");
//true

What gives?

I've tested it with multiple regex and it seems the /g flag makes all ! 's return true.

I'm using Chrome 34 if that helps.

Answer 1

From the MDN page on .test() :

Use test() whenever you want to know whether a pattern is found in a string (similar to the String.search method); for more information (but slower execution) use the exec method (similar to the String.match method). As with exec (or in combination with it), test called multiple times on the same global regular expression instance will advance past the previous match.

So, using the g flag will move through the string on each successive call to .test() advancing past each match it finds (same behavior as .exec() ).

It does this by modifying a property of the regex itself called lastIndex which tells the next .test() operation where it should start searching in the string the next time the regex is used (with certain methods). It is the g flag that tells the regex to use the .lastIndex property. Without that flag, that property is not used. That property can be manually set back to 0 , but if you just remove the g flag, then you don't have to worry about it.

Unless you are explicitly trying to use this capability, you probably don't want to use the g flag with .test() .

Answer 2

The first time you call re2.test(), it move a pointer index property called lastIndex to the end of pattern string of re2 object. The second time you call it, it will return false because it can not find any match of "abc" after "c" character of pattern string

After each calling .test(), you can reset it by re2.lastIndex = 0 . So the second time you call .test(), it will search from first index

This property only works if the "g" modifier is set.