QSemaphore: Fix 64-bit Linux QSemaphores initialized to non-zero

The original design for the 64-bit futex had the token count replicated
in the high part. But the constructor wasn't setting it. This is one of
the issues I had noticed when investigating QTBUG-66875, but didn't need
to address since the the fix I ended up applying in commit
081c001deb made that unnecessary: the high
part only had the number of waiters.

Unfortunately, when commit f502270c0f06daba75c1da381bd1658d81aa7bba
brought back the token count in the high part, I failed to correct that
problem. As a consequence, every QSemaphore that was initialized in the
constructor to a non-zero value would eventually deadlock.

This commit fixes that oversight.

Task-number: QTBUG-67481
Change-Id: I662b8f168c74440ab1a8fffd1522be6b85adb4d0
Reviewed-by: Lars Knoll <lars.knoll@qt.io>
Reviewed-by: Mårten Nordheim <marten.nordheim@qt.io>
Reviewed-by: Svenn-Arne Dragly <svenn-arne.dragly@qt.io>

src/corelib/thread/qsemaphore.cpp

@@ -291,11 +291,15 @@ public:
 QSemaphore::QSemaphore(int n)
 {
     Q_ASSERT_X(n >= 0, "QSemaphore", "parameter 'n' must be non-negative");
-    if (futexAvailable())
-        u.store(n);
-    else
+    if (futexAvailable()) {
+        quintptr nn = unsigned(n);
+        if (futexHasWaiterCount)
+            nn |= quint64(nn) << 32;    // token count replicated in high word
+        u.store(nn);
+    } else {
         d = new QSemaphorePrivate(n);
     }
+}
 
 /*!
     Destroys the semaphore.