Try contextually converting condition of constexpr if to Boolean value

Summary:
C++1z 6.4.1/p2:
 If the if statement is of the form if constexpr, the value of the
 condition shall be a contextually converted constant expression of type
 bool [...]
C++1z 5.20/p4:
 [...] A contextually converted constant expression of type bool is an
 expression, contextually converted to bool (Clause4), where the
 converted expression is a constant expression and the conversion
 sequence contains only the conversions above. [...]

Contextually converting result of an expression `e` to a Boolean value
requires `bool t(e)` to be well-formed.

An explicit conversion function is only considered as a user-defined
conversion for direct-initialization, which is essentially what
//contextually converted to bool// requires.

Also, fixes PR28470.

Reviewers: rsmith

Subscribers: cfe-commits

Differential Revision: https://reviews.llvm.org/D24158

llvm-svn: 280838

1 files changed, 13 insertions, 0 deletions

diff --git a/clang/test/CodeGenCXX/cxx1z-constexpr-if.cpp b/clang/test/CodeGenCXX/cxx1z-constexpr-if.cpp
index 80a397f51e9..14695363b33 100644
--- a/clang/test/CodeGenCXX/cxx1z-constexpr-if.cpp
+++ b/clang/test/CodeGenCXX/cxx1z-constexpr-if.cpp
@@ -2,7 +2,15 @@
 
 void should_be_used_1();
 void should_be_used_2();
+void should_be_used_3();
 void should_not_be_used();
+
+struct A {
+  constexpr explicit operator bool() const {
+    return true;
+  }
+};
+
 void f() {
   if constexpr (false)
     should_not_be_used();
@@ -15,7 +23,12 @@ void f() {
     goto foo;
 foo: should_not_be_used();
   }
+  if constexpr (A())
+    should_be_used_3();
+  else
+    should_not_be_used();
 }
 
 // CHECK: should_be_used_1
 // CHECK: should_be_used_2
+// CHECK: should_be_used_3