Change Expr::isLvalue() to allow the "void" type. This fixes bz2000 submitted by Neil Booth.

Neil, can you point me to the place in the C99 spec that says this is allowed? I thought Expr::isLvalue() conformed to the spec, which says "C99 6.3.2.1: an lvalue is an expression with an object type or an incomplete type other than void.". Please advise.



git-svn-id: https://llvm.org/svn/llvm-project/cfe/trunk@46917 91177308-0d34-0410-b5e6-96231b3b80d8

2 files changed, 7 insertions, 4 deletions

diff --git a/AST/Expr.cpp b/AST/Expr.cpp
index f02e817f06..f979c7aa7b 100644
--- a/AST/Expr.cpp
+++ b/AST/Expr.cpp
@@ -357,9 +357,6 @@ Expr::isLvalueResult Expr::isLvalue() const {
   if (TR->isFunctionType()) // from isObjectType()
     return LV_NotObjectType;
 
-  if (TR->isVoidType())
-    return LV_IncompleteVoidType;
-
   if (TR->isReferenceType()) // C++ [expr]
     return LV_Valid;
 
diff --git a/test/Sema/deref.c b/test/Sema/deref.c
index 7441584107..7efb55e720 100644
--- a/test/Sema/deref.c
+++ b/test/Sema/deref.c
@@ -17,6 +17,12 @@ void foo2 (void)
 void foo3 (void)
 {
  void* x = 0;
- void* y = &*x; // expected-error {{address expression must be an lvalue or a function designator}}
+ void* y = &*x;
+}
+
+extern const void cv1;
+const void *foo4 (void)
+{
+  return &cv1;
 }