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Malloc should return address aligned to 8 bytes #100

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Merged
merged 1 commit into from
Jan 4, 2024
Merged

Malloc should return address aligned to 8 bytes #100

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10 changes: 3 additions & 7 deletions test/malloc_compliance_tests.cpp
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Original file line number Diff line number Diff line change
Expand Up @@ -18,6 +18,8 @@
#include "base.hpp"
using umf_test::test;

#define ALLOC_MIN_ALIGNMENT 8

//------------------------------------------------------------------------
// Configurable defs
//------------------------------------------------------------------------
Expand Down Expand Up @@ -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)
Expand Down