Commit 7eda3368 by Jason Merrill Committed by Jason Merrill

re PR c++/48289 (-pedantic breaks std::move)

	PR c++/48289
	* pt.c (build_non_dependent_expr): Keep dereferences outside the
	NON_DEPENDENT_EXPR.

From-SVN: r171461
parent c7dc8804
2011-03-25 Jason Merrill <jason@redhat.com>
PR c++/48289
* pt.c (build_non_dependent_expr): Keep dereferences outside the
NON_DEPENDENT_EXPR.
2011-03-25 Kai Tietz <ktietz@redhat.com>
* decl.c (decls_match): Replace target hook
......
......@@ -18846,24 +18846,17 @@ build_non_dependent_expr (tree expr)
TREE_OPERAND (expr, 0),
build_non_dependent_expr (TREE_OPERAND (expr, 1)));
/* Keep dereferences outside the NON_DEPENDENT_EXPR so lvalue_kind
doesn't need to look inside. */
if (TREE_CODE (expr) == INDIRECT_REF && REFERENCE_REF_P (expr))
return convert_from_reference (build_non_dependent_expr
(TREE_OPERAND (expr, 0)));
/* If the type is unknown, it can't really be non-dependent */
gcc_assert (TREE_TYPE (expr) != unknown_type_node);
/* Otherwise, build a NON_DEPENDENT_EXPR.
REFERENCE_TYPEs are not stripped for expressions in templates
because doing so would play havoc with mangling. Consider, for
example:
template <typename T> void f<T& g>() { g(); }
In the body of "f", the expression for "g" will have
REFERENCE_TYPE, even though the standard says that it should
not. The reason is that we must preserve the syntactic form of
the expression so that mangling (say) "f<g>" inside the body of
"f" works out correctly. Therefore, the REFERENCE_TYPE is
stripped here. */
return build1 (NON_DEPENDENT_EXPR, non_reference (TREE_TYPE (expr)), expr);
/* Otherwise, build a NON_DEPENDENT_EXPR. */
return build1 (NON_DEPENDENT_EXPR, TREE_TYPE (expr), expr);
}
/* ARGS is a vector of expressions as arguments to a function call.
......
2011-03-25 Jason Merrill <jason@redhat.com>
* g++.dg/cpp0x/move1.C: New.
2011-03-25 Ira Rosen <ira.rosen@linaro.org>
PR target/48287
......
// { dg-options "-std=c++0x -pedantic-errors" }
#include <utility>
class A { };
static void g ( A && ) { }
template < class T > class B {
public:
void f ( ) {
A a;
g ( std :: move ( a ) );
}
};
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