Question

>>> is lexed as >> >. But what happens if the first > closes a template argument list, should the result be equivalent to > > > or > >>?

It does matter in the following code:

template<class T> struct X { };

void operator >>(const X<int>&, int) { }

int main() {
    *new X<int>>> 1;
}

Answer 1

The text of the FDIS says

Similarly, the first non-nested >> is treated as two consecutive but distinct > tokens

It cannot unlex tokens and relex. So this will be a > > >. Note that the input to a C++ implementation is first lexed into preprocessing tokens, and then those tokens are converted into C++ tokens. So first your input are the C++ tokens >> >, then the C++ parser changes these to > > >.

Each preprocessing token is converted into a token. (2.7). The resulting tokens are syntactically and semantically analyzed and translated as a translation unit. [ Note: The process of analyzing and translating the tokens may occasionally result in one token being replaced by a sequence of other tokens (14.2). — end note ]

There’s no chance you could merge those two trailing > > tokens.