I could not understand those 3 rules in the specs too well -- hope to have something that is more plain English -- but here is what I gathered from JavaScript: The Definitive Guide, 6th Edition, David Flanagan, O'Reilly, 2011:

Quote:

JavaScript does not treat every line break as a semicolon: it usually treats line breaks as semicolons only if it can’t parse the code without the semicolons.

Another quote: for the code

var a a = 3 console.log(a)

JavaScript does not treat the second line break as a semicolon because it can continue parsing the longer statement a = 3;

and:

two exceptions to the general rule that JavaScript interprets line breaks as semicolons when it cannot parse the second line as a continuation of the statement on the first line. The first exception involves the return, break, and continue statements ... If a line break appears after any of these words ... JavaScript will always interpret that line break as a semicolon. ... The second exception involves the ++ and −− operators ... If you want to use either of these operators as postfix operators, they must appear on the same line as the expression they apply to. Otherwise, the line break will be treated as a semicolon, and the ++ or -- will be parsed as a prefix operator applied to the code that follows. Consider this code, for example:

x ++ y

It is parsed as x; ++y; , not as x++; y

So I think to simplify it, that means:

In general, JavaScript will treat it as continuation of code as long as it makes sense -- except 2 cases: (1) after some keywords like return , break , continue , and (2) if it sees ++ or -- on a new line, then it will add the ; at the end of the previous line.

The part about "treat it as continuation of code as long as it makes sense" makes it feel like regular expression's greedy matching.

With the above said, that means for return with a line break, the JavaScript interpreter will insert a ;

(quoted again: If a line break appears after any of these words [such as return ] ... JavaScript will always interpret that line break as a semicolon)

and due to this reason, the classic example of

return { foo: 1 }

will not work as expected, because the JavaScript interpreter will treat it as:

return; // returning nothing { foo: 1 }

There has to be no line-break immediately after the return :

return { foo: 1 }

for it to work properly. And you may insert a ; yourself if you were to follow the rule of using a ; after any statement: