Is [\s\S] same as . (dot)?

Question

When we include shorthand for character class and negated-character class in same character class, is it same as dot . which mean any character ?

I did a test on regex101.com and every character matched.

Is [\s\S] [\w\W] and [\d\D] same as . ?

I want to know if this behavior is persistent in web's front and backend languages like Javascript, Php, Python and others.

Answer 1

"No" it is not the same. It has an important difference if you are not using the single line flag (meaning that . does not match all).

The [\s\S] comes handy when you want to do mix of matches when the . does not match all.

It is easier to explain it with an example. Suppose you want to capture whatever is between a and b, so you can use pattern a(.*?)b (? is for ungreedy matches and parentheses for capturing the content), but if there are new lines suppose you don't want to capture this in the same group, so you can have another regex like a([\s\S]*?)b.

Therefore if we create one pattern using both approaches it results in:

a(.*)b|a([\s\S]*?)b

In this case, if you see the scenario in regex101, then you will have a colorful and easy way to differentiate the scenarios (in green capturing group #1 and in red capturing group #2):

So, in conclusion, the [\s\S] is a regex trick when you want to match multiple lines and the . does not suit your needs. It basically depends on your use case.

However, if you use the single line flag where . matches new lines, then you don't need the regex trick, below you can see that all is green and group 2 (red above) is not matched:

Have also created a javascript performance test and it impacts in the performance around 25%:

https://jsperf.com/ss-vs-dot

Answer 2

The answer is: It depends.
If your regex engine does match every character with . then yes, the result is the same. If it doesn't then no, the result is not the same. In standard JavaScript . , for example, does not match line breaks.