[Lldb-commits] [lldb] [lldb/docs] Fix/improve the gdb command map for dynamic types (PR #138538)

[Pavel Labath via lldb-commits]

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@@ -800,16 +800,24 @@ Print the dynamic type of the result of an expression
   (gdb) p someCPPObjectPtrOrReference
   (Only works for C++ objects)
 
+LLDB does this automatically if determining the dynamic type does not require
+running the target (in C++, running the target is never needed). This default is
+controlled by the `target.prefer-dynamic-value` setting. If that is disabled, it
+can be re-enabled on a per-command basis:
+
 .. code-block:: shell
 
-  (lldb) expr -d 1 -- [SomeClass returnAnObject]
-  (lldb) expr -d 1 -- someCPPObjectPtrOrReference
+  (lldb) settings set target.prefer-dynamic-value no-dynamic-values
+  (lldb) frame variable -d no-run-target someCPPObjectPtrOrReference
+  (lldb) expr -d no-run-target -- someCPPObjectPtr
 
-or set dynamic type printing to be the default:
+Note that printing of the dynamic type of references is not possible with the
+`expr` command. The workaround is to take the address of the reference and
+instruct lldb to print the children of the resulting pointer.
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labath wrote:

(If you're wondering why we're making a copy of a reference -- but not of a pointer -- I suspect that's because in the expression evaluator we do not differentiate between an object and a reference to it. I suppose we could do it, and then somehow treat references as pointers, but I don't exactly know what would that entail.)

https://github.com/llvm/llvm-project/pull/138538