cppcheck confuses template operator() call for "less than" comparison

[VG]

The following code triggers an error: "AST broken, binary operator '>' doesn't have two operands." This is obviously because the expression has been parsed incorrectly.

template <std::size_t First, std::size_t... Indices, typename Functor>
constexpr void constexpr_for_fold_impl([[maybe_unused]] Functor&& f, std::index_sequence<Indices...>) noexcept
{
    (std::forward<Functor>(f).template operator() < First + Indices > (), ...);
}

[Paul Fultz]

So cppcheck removes the template keyword which makes it harder to know what brackets should be linked. We should add a template flag in the token when its removed so we can later find templates.

Alternatively, I wonder if we can try to link the brackets after the AST is built. Since we know there is no AST for the < it must mean it could be a template bracket.

[VG]

I think the 'template' keyword here is absolutely necessary for disambiguation and you can't parse this correctly if it was removed (i. e. without taking this keyword into account).

[Paul Fultz]

Well, actually we dont remove the template keyword in the snippet above. However, we still dont check for a template keyword before ahead anyways. So I fixed this here:

https://github.com/danmar/cppcheck/pull/3399

It will remove the template keyword and then set the isTemplate flag on the name that follows, and then linking the brackets will use this flag.