Don't copy the functor object for each signal emission

The behavior was different in the variadic template code and in the C++98
code. The code without variadic template support was not copying the functor
object (e.g. a lambda) before calling it.
However, in the variadic template section, QtPrivate::FunctorCall::call
took the functor by value instead of by reference resulting in a copy.

QtPrivate::FunctorCall::call is a helper function for
QtPrivate::FunctionPointer::call which is only needed for variadic template
expension.

[ChangeLog][QtCore][QObject] If the compiler supports variadic templates,
no longer copy functor connected to a signal each time the signal is
emitted. Restoring the C++98 behavior.

Task-number: QTBUG-52542
Change-Id: I3ca20ef6910893b8a288e70af7de4c7b69502173
Reviewed-by: Thiago Macieira <thiago.macieira@intel.com>

1 files changed, 1 insertions, 1 deletions

diff --git a/src/corelib/kernel/qobjectdefs_impl.h b/src/corelib/kernel/qobjectdefs_impl.h
index d5574a4dd1..922b6bbb42 100644
--- a/src/corelib/kernel/qobjectdefs_impl.h
+++ b/src/corelib/kernel/qobjectdefs_impl.h
@@ -491,7 +491,7 @@ namespace QtPrivate {
     template <typename, typename, typename, typename> struct FunctorCall;
     template <int... II, typename... SignalArgs, typename R, typename Function>
     struct FunctorCall<IndexesList<II...>, List<SignalArgs...>, R, Function> {
-        static void call(Function f, void **arg) {
+        static void call(Function &f, void **arg) {
             f((*reinterpret_cast<typename RemoveRef<SignalArgs>::Type *>(arg[II+1]))...), ApplyReturnValue<R>(arg[0]);
         }
     };