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PR18746: If a constexpr function has a dependent return type and no return 

statements, don't diagnose; the return type might end up being 'void'.

Patch by Rahul Jain! Tiny tweaks by me.

llvm-svn: 206929
This commit is contained in:
Richard Smith 2014-04-22 23:14:23 +00:00
parent 637cac42ed
commit 06ffb45ce4
2 changed files with 11 additions and 2 deletions
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6
clang/lib/Sema/SemaDeclCXX.cpp
View File

@ -1175,10 +1175,12 @@ bool Sema::CheckConstexprFunctionBody(const FunctionDecl *Dcl, Stmt *Body) {
} else {
if (ReturnStmts.empty()) {
// C++1y doesn't require constexpr functions to contain a 'return'
// statement. We still do, unless the return type is void, because
// statement. We still do, unless the return type might be void, because
// otherwise if there's no return statement, the function cannot
// be used in a core constant expression.
bool OK = getLangOpts().CPlusPlus1y && Dcl->getReturnType()->isVoidType();
bool OK = getLangOpts().CPlusPlus1y &&
(Dcl->getReturnType()->isVoidType() ||
Dcl->getReturnType()->isDependentType());
Diag(Dcl->getLocation(),
OK ? diag::warn_cxx11_compat_constexpr_body_no_return
: diag::err_constexpr_body_no_return);

7
clang/test/SemaCXX/cxx1y-deduced-return-type.cpp
View File

@ -261,6 +261,13 @@ namespace DefaultedMethods {
namespace Constexpr {
constexpr auto f1(int n) { return n; }
template<typename T> struct X { constexpr auto f() {} }; // PR18746
template<typename T> struct Y { constexpr T f() {} }; // expected-note {{control reached end of constexpr function}}
void f() {
X<int>().f();
Y<void>().f();
constexpr int q = Y<int>().f(); // expected-error {{must be initialized by a constant expression}} expected-note {{in call to '&Y<int>()->f()'}}
}
struct NonLiteral { ~NonLiteral(); } nl; // expected-note {{user-provided destructor}}
constexpr auto f2(int n) { return nl; } // expected-error {{return type 'Constexpr::NonLiteral' is not a literal type}}
}