[Lldb-commits] [lldb] [lldb] Override default struct layout when building register types (PR #189590)

[Sergei Barannikov via lldb-commits]

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@@ -28,31 +28,18 @@ static void dump_type_value(lldb_private::CompilerType &fields_type, T value,
                             lldb_private::Stream &strm) {
   lldb::ByteOrder target_order = exe_scope->CalculateProcess()->GetByteOrder();
 
-  // For the bitfield types we generate, it is expected that the fields are
-  // in what is usually a big endian order. Most significant field first.
-  // This is also clang's internal ordering and the order we want to print
-  // them. On a big endian host this all matches up, for a little endian
-  // host we have to swap the order of the fields before display.
-  if (target_order == lldb::ByteOrder::eByteOrderLittle) {
-    value = reg_info.flags_type->ReverseFieldOrder(value);
-  }
-
-  // Then we need to match the target's endian on a byte level as well.
+  // The type will be rendered in the target's type system, so it must match
+  // its endian.
   if (lldb_private::endian::InlHostByteOrder() != target_order)
     value = llvm::byteswap(value);
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s-barannikov wrote:

> Yes that's a good find but the class `RegisterValue` can in theory hold other endian.

Can it useful somehow? My gut feeling is that it will only make the code even more convoluted and we should stick to host endian (at least in RegisterValue).

However, it appears more complicated that I imagined, so I'll just resolve this comment and dig into it more so that I continue the discussion.


https://github.com/llvm/llvm-project/pull/189590