Fix USE rename when use-name and local-name are the same.
Previously, the associated symbol was being removed from scope.
Operator rename implementation needed no change, because, as it
doesn't call AddUseRename(), symbol erasure is skipped.
Fixes #60223
Reviewed By: klausler
Differential Revision: https://reviews.llvm.org/D143933
const auto &useName{std::get<1>(x.t)};
AddUseRename(useName.source);
SymbolRename rename{AddUse(localName.source, useName.source)};
- if (rename.use) {
+ if (rename.use && localName.source != useName.source) {
EraseRenamedSymbol(*rename.use);
}
Resolve(useName, rename.use);
--- /dev/null
+! RUN: %python %S/test_errors.py %s %flang_fc1
+! Test rename to the same name.
+module m1
+ integer, allocatable :: a(:)
+
+ interface operator(.add.)
+ module procedure add
+ end interface
+
+contains
+ integer function add(a, b)
+ integer, intent(in) :: a, b
+
+ add = a + b
+ end function
+end
+
+program p1
+ use m1, a => a, operator(.add.) => operator(.add.)
+
+ allocate(a(10))
+ deallocate(a)
+ print *, 2 .add. 2
+end