`name` depends on the template type, so we have to explicitly say
that `has` is a template with the `template` keyword.
Rather than do that I'm putting back the explicit type for name.
Original-commit: flang-compiler/f18@
0cde87f2fe64801dbbd2172970e3d0113d421006
Reviewed-on: https://github.com/flang-compiler/f18/pull/65
if (!funcRef) {
return;
}
- auto *name = std::get_if<parser::Name>(
+ parser::Name *name = std::get_if<parser::Name>(
&std::get<parser::ProcedureDesignator>((*funcRef)->v.t).u);
if (!name || !name->symbol || !name->symbol->has<EntityDetails>()) {
return;