It's a regular expression literal, similar to using return new RegExp('^(\d{4}|\d{6})$').test(pin) The literal part implies that it's a means of representing a specific data type as a string in code—just like true and 'true' are different, as one is a boolean literal and the other is a string literal.

Specifically, the regex ^(d{4}|d{6})$ breaks down to:

^       a string that starts with...

(       either

  d    a digit (0-9)...

  {4}   that repeats four times...

|       or

  d    a digit (0-9)...

  {6}   that repeats six times...

)

$       and then ends

So: '1234', '123456', etc would match. '123.00', '12345','abc123',' 1234', ' 1234 ' would not match.

As noted by several others in the comments on Draco18s' answer there are several nuances to be aware of with using regex literals in JS:

• The literal syntax doesn't require you to escape special characters within the regex pattern. Using the RegExp constructor requires you to represent the pattern as a string, which in turn requires escaping. Note the differences of the 's between the two syntaxes.




• Using a regex literal will treat the regex as a constant, whereas using new RegExp() leaves life cycle management of the regex instance up to you.

  
  
  

  
  

  The literal notation is compiled and implies a constant regex, whereas the constructor version is reparsed from the string, and so the literal is better optimized/cached. jsperf.com/regexp-literal-vs-constructor/4 Note: you can get basically the same effect by caching the new Regex in a variable, but the literal one is cached at the JIT step – user120242

  
  

  
  
  

  In other words, using a regex literal can avoid potential performance pitfalls:

  
  
  

  Example:

  
  
  

  for (var i = 0; i < 1000; i++) {
  
    // Instantiates 1x Regex per iteration
  
    var constructed = new RegExp('^(\d{4}|\d{6})$') 
  
  
  
    // Instantiates 1 Regex
  
    var literal = /^(d{4}|d{6})$/ 
  
  }