Malloc should return address aligned to 8 bytes

Ref: #79

Signed-off-by: Lukasz Dorau <[email protected]>

Author: ldorau

test/malloc_compliance_tests.cpp

@@ -18,6 +18,8 @@
 #include "base.hpp"
 using umf_test::test;
 
+#define ALLOC_MIN_ALIGNMENT 8
+
 //------------------------------------------------------------------------
 // Configurable defs
 //------------------------------------------------------------------------
@@ -57,15 +59,9 @@ void malloc_compliance_test(umf_memory_pool_handle_t hPool) {
         alloc_ptr_size[i] = rand_alloc_size(MAX_ALLOC_SIZE);
         alloc_ptr[i] = umfPoolMalloc(hPool, alloc_ptr_size[i]);
 
-#ifndef _WIN32 // TODO: explain why it fails on some Windows systems
-        // Supported under USE_WEAK_ALIGNMENT_RULES macro
-        // For compatibility, on 64-bit systems malloc should align to 16 bytes
-        size_t alignment =
-            (sizeof(intptr_t) > 4 && alloc_ptr_size[i] > 8) ? 16 : 8;
-        ASSERT_NE(addressIsAligned(alloc_ptr[i], alignment), 0)
+        ASSERT_NE(addressIsAligned(alloc_ptr[i], ALLOC_MIN_ALIGNMENT), 0)
             << "Malloc should return pointer that is suitably aligned so that "
                "it may be assigned to a pointer to any type of object";
-#endif
 
         // Fill memory with a pattern ((id of the allocation) % 256)
         ASSERT_NE(alloc_ptr[i], nullptr)