[Lldb-commits] [lldb] [lldb] Print empty enums as if they were unrecognised normal enums (PR #97553)

[David Spickett via lldb-commits]

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@@ -8649,19 +8649,24 @@ static bool DumpEnumValue(const clang::QualType &qual_type, Stream &s,
   // At the same time, we're applying a heuristic to determine whether we want
   // to print this enum as a bitfield. We're likely dealing with a bitfield if
   // every enumerator is either a one bit value or a superset of the previous
-  // enumerators. Also 0 doesn't make sense when the enumerators are used as
-  // flags.
-  for (auto *enumerator : enum_decl->enumerators()) {
-    uint64_t val = enumerator->getInitVal().getSExtValue();
-    val = llvm::SignExtend64(val, 8*byte_size);
-    if (llvm::popcount(val) != 1 && (val & ~covered_bits) != 0)
-      can_be_bitfield = false;
-    covered_bits |= val;
-    ++num_enumerators;
-    if (val == enum_svalue) {
-      // Found an exact match, that's all we need to do.
-      s.PutCString(enumerator->getNameAsString());
-      return true;
+  // enumerators. Also an enumerator of 0 doesn't make sense when the
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DavidSpickett wrote:

I clarified this to "an enumerator of 0 doesn't make sense", which is my understanding of it. As opposed to "a value of 0 doesn't make sense".

https://github.com/llvm/llvm-project/pull/97553