Atomics: remove requirement for alignment equality with plain types

This was originally added so that you could replace a T with
QAtomicInteger<T> in the same class and still keep ABI. However, for
legacy reasons, on 32-bit x86, types larger than 4 bytes keep an old
1990s alignment of only 4 bytes, but modern std::atomic<T> for those 8-
byte types enforces an alignment of 8 bytes. Therefore, the requirement
to keep alignment is not possible to guarantee.

In other words: you may not replace T with QAtomicInteger<T> or
std::atomic<T> and assume no ABI breakages in all platforms.

This is a requirement to implement atomicity. An 8-byte type aligned to
only a 4-byte boundary could cross a 16-byte boundary or, worse, cross a
cacheline boundary. Crossing the 16-byte boundary could be bad on some
processors, but crossing the cacheline boundary (addresses ending in
0x3C, 0x7C, 0xCC and 0xFC, or 4 out of 64 possible addresses or 6.25%)
is always bad: the CPUs cannot guarantee an atomic load or store
operation.

See also <https://gcc.gnu.org/bugzilla/show_bug.cgi?id=71660>.

Task-number: QTBUG-67858
Change-Id: If90a92b041d3442fa0a4fffd15283e4615474582
Reviewed-by: Frederik Gladhorn <frederik.gladhorn@qt.io>

tests/auto/corelib/thread/qatomicinteger/tst_qatomicinteger.cpp

@@ -185,12 +185,10 @@ private Q_SLOTS:
 };
 
 template <bool> inline void booleanHelper() { }
-struct TypeInStruct { TEST_TYPE type; };
 
 void tst_QAtomicIntegerXX::static_checks()
 {
     Q_STATIC_ASSERT(sizeof(QAtomicInteger<T>) == sizeof(T));
-    Q_STATIC_ASSERT(Q_ALIGNOF(QAtomicInteger<T>) == Q_ALIGNOF(TypeInStruct));
 
     // statements with no effect
     (void) QAtomicInteger<T>::isReferenceCountingNative();