Make scalar() propagate lvalueness

As mentioned in ticket #24346, scalar() should not change lvalueness.
The fact that it did was a side effect of the implementation and a
bug.  foo(scalar substr(....)) should pass a substr lvalue to foo just
as it would without scalar() or with a $ prototype (which is meant to
be equivalent to scalar()).

This also makes it possible to force scalar context in list assignment
to lvalue subroutines, as in (foo(), scalar bar()) = @list.

1 files changed, 8 insertions, 1 deletions

diff --git a/t/op/substr.t b/t/op/substr.t
index ceacdf6e47..abd5d7f630 100644
--- a/t/op/substr.t
+++ b/t/op/substr.t
@@ -23,7 +23,7 @@ $SIG{__WARN__} = sub {
 
 BEGIN { require './test.pl'; }
 
-plan(381);
+plan(382);
 
 run_tests() unless caller;
 
@@ -684,6 +684,13 @@ is($x, "\x{100}\x{200}\xFFb");
     }
 }
 
+# Also part of perl #24346; scalar(substr...) should not affect lvalueness
+{
+    my $str = "abcdef";
+    sub { $_[0] = 'dea' }->( scalar substr $str, 3, 2 );
+    is $str, 'abcdeaf', 'scalar does not affect lvalueness of substr';
+}
+
 # [perl #24200] string corruption with lvalue sub
 
 {