Use 8-byte alignment for FixedSizeAllocatorWithCleanup when 16-byte alignment is false (GH #992)

1 files changed, 14 insertions, 4 deletions

diff --git a/secblock.h b/secblock.h
index 9be03050..39beb677 100644
--- a/secblock.h
+++ b/secblock.h
@@ -547,9 +547,14 @@ private:
 

 #else

 

-	// CRYPTOPP_BOOL_ALIGN16 is 0. Use natural alignment of T.

+	// CRYPTOPP_BOOL_ALIGN16 is 0. Normally we would use the natural

+	// alignment of T. The problem we are having is, some toolchains

+	// are changing the boundary for 64-bit arrays. 64-bit elements

+	// require 8-byte alignment, but the toolchain is laying the array

+	// out on a 4 byte boundary. See GH #992 for mystery alignment:

+	// https://github.com/weidai11/cryptopp/issues/992

 	T* GetAlignedArray() {return m_array;}

-	T m_array[S];

+	CRYPTOPP_ALIGN_DATA(8) T m_array[S];

 

 #endif

 

@@ -700,9 +705,14 @@ public:
 

 private:

 

-	// T_Align16 is false. Use natural alignment of T.

+	// T_Align16 is false. Normally we would use the natural

+	// alignment of T. The problem we are having is, some toolchains

+	// are changing the boundary for 64-bit arrays. 64-bit elements

+	// require 8-byte alignment, but the toolchain is laying the array

+	// out on a 4 byte boundary. See GH #992 for mystery alignment:

+	// https://github.com/weidai11/cryptopp/issues/992

 	T* GetAlignedArray() {return m_array;}

-	T m_array[S];

+	CRYPTOPP_ALIGN_DATA(8) T m_array[S];

 

 	A m_fallbackAllocator;

 	bool m_allocated;