Why can’t you treat a FILETIME as an __int64?

Raymond Chen

The FILETIME structure represents a 64-bit value in two parts:

typedef struct _FILETIME {
  DWORD dwLowDateTime;
  DWORD dwHighDateTime;

You may be tempted to take the entire FILETIME structure and access it directly as if it were an __int64. After all, its memory layout exactly matches that of a 64-bit (little-endian) integer. Some people have written sample code that does exactly this:

pi = (__int64*)&ft; // WRONG
(*pi) += (__int64)num*datepart; // WRONG

Why is this wrong?


Since a FILETIME is a structure containing two DWORDs, it requires only 4-byte alignment, since that is sufficient to put each DWORD on a valid DWORD boundary. There is no need for the first DWORD to reside on an 8-byte boundary. And in fact, you’ve probably already used a structure where it doesn’t: The WIN32_FIND_DATA structure.

typedef struct _WIN32_FIND_DATA {
    DWORD dwFileAttributes;
    FILETIME ftCreationTime;
    FILETIME ftLastAccessTime;
    FILETIME ftLastWriteTime;
    DWORD nFileSizeHigh;
    DWORD nFileSizeLow;
    DWORD dwReserved0;
    DWORD dwReserved1;
    TCHAR  cFileName[ MAX_PATH ];
    TCHAR  cAlternateFileName[ 14 ];

Observe that the three FILETIME structures appear at offsets 4, 12, and 20 from the beginning of the structure. They have been thrown off 8-byte alignment by the dwFileAttributes member.

Casting a FILETIME to an __int64 therefore can (and in the WIN32_FIND_DATA case, will) create a misaligned pointer. Accessing a misaligned pointer will raise a STATUS_DATATYPE_MISALIGNMENT exception on architectures which require alignment.

Even if you are on a forgiving platform that performs automatic alignment fixups, you can still run into trouble. More on this and other consequences of alignment in the next few entries.

Exercise: Why are the LARGE_INTEGER and ULARGE_INTEGER structures not affected?


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