================
@@ -27,12 +27,23 @@ struct E {
constexpr E(){};
} TestE;
-// Declared but not defined constexpr constructor should not emit full debug
info..
-// CHECK-DAG: !DICompositeType(tag: DW_TAG_structure_type, name:
"DeclaredConstexpr"{{.*}}flags: DIFlagFwdDecl
+// A constexpr constructor that is only declared here may still be defined
+// elsewhere and called, so it cannot be relied on to home the type.
+// CHECK-DAG: !DICompositeType(tag: DW_TAG_structure_type, name:
"DeclaredConstexpr"{{.*}}DIFlagTypePassByValue
struct DeclaredConstexpr {
constexpr DeclaredConstexpr();
} TestDeclaredConstexpr;
+// A declared-only constexpr constructor that is never odr-used need not be
+// defined anywhere in the program, so nothing would ever emit the definition
+// of the type - but the type is still required to be complete here.
+// CHECK-DAG: !DICompositeType(tag: DW_TAG_structure_type, name:
"ConstexprDeclaredOnly"{{.*}}DIFlagTypePassByValue
+struct ConstexprDeclaredOnly {
+ unsigned long v;
+ constexpr ConstexprDeclaredOnly(unsigned long t);
----------------
ClaytonKnittel wrote:
If you declare a constexpr constructor, then in order to construct this type,
you must invoke this constructor, right? Even if you call it from a
non-constexpr context, the program won't link unless the constructor is defined
somewhere.
There is no context that I can think of that would allow you to construct this
type through this constructor without being able to see the constructors
definition (which would opt the TU out of ctor homing), and without the
constructor defined elsewhere (which would opt the defining TU out of ctor
homing).
The method below, in a real program, could never be invoked, unless you do
something like `reinterpret_cast` memory constructed as some other type to
`ConstexprDeclaredOnly`, which is UB (by my understanding).
https://github.com/llvm/llvm-project/pull/221566
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