c++: Fix return deduction of lambda in discarded stmt.

A return statement in a discarded statement is not used for return type
deduction, but we still want to do deduction for a return statement in a
lambda in a discarded statement.

	PR c++/93442
	* parser.c (cp_parser_lambda_expression): Clear in_discarded_stmt.

gcc/cp/ChangeLog

 2020-01-28  Jason Merrill  <jason@redhat.com>
 
+	PR c++/93442
+	* parser.c (cp_parser_lambda_expression): Clear in_discarded_stmt.
+
 	PR c++/93477
 	PR c++/91476
 	* decl2.c (copy_linkage): Do copy DECL_ONE_ONLY and DECL_WEAK.

gcc/cp/parser.c

@@ -10530,6 +10530,10 @@ cp_parser_lambda_expression (cp_parser* parser)
     parser->implicit_template_scope = 0;
     parser->auto_is_implicit_function_template_parm_p = false;
 
+    /* The body of a lambda in a discarded statement is not discarded.  */
+    bool discarded = in_discarded_stmt;
+    in_discarded_stmt = 0;
+
     /* By virtue of defining a local class, a lambda expression has access to
        the private variables of enclosing classes.  */
 
@@ -10560,6 +10564,8 @@ cp_parser_lambda_expression (cp_parser* parser)
 
     finish_struct (type, /*attributes=*/NULL_TREE);
 
+    in_discarded_stmt = discarded;
+
     parser->num_template_parameter_lists = saved_num_template_parameter_lists;
     parser->in_statement = in_statement;
     parser->in_switch_statement_p = in_switch_statement_p;

gcc/testsuite/g++.dg/cpp1z/constexpr-if-lambda1.C

+// PR c++/93442
+// { dg-do compile { target c++17 } }
+
+struct bar {
+    int foo(){return 0;}
+};
+
+int foobar() {
+    if constexpr(true) {
+        return 0;
+    } else {
+        return [](){
+            return bar{};
+        }().foo();
+    }
+}