DWORD64 if set to -1 gives 32 1s i.e. 0xffffffffffffffffffffffffffffffff

Question

C++ code (Visual studio started with devenv /useenv (x64) and isWOW64 is false)

DWORD64 check;

check = -1;
printf("value %u", check);

it prints the value 4294967295 i.e. 0x(32)f which is the same if i do it with simple DWORD in an x32 environment

yes i know DWORD64 is unsigned __int64, but shouldn't it be 0x(64)f ?

what did the assembler do there ? disassembling the code didn't help me much.

Answer 1

You problem here lies in the printf format string. Using %u tells printf to print a 32 bits value. Hence, it uses only the first 32 bits of your DWORD64. To print all the 64 bits that has been pushed onto the stack, use %llu (for unsigned long long).

Note also that DWORD64 cannot be unsigned long. On all Windows versions, even on Windows 64 bits, long is 32 bits.

See LLP64 model.

Answer 2

If you look at the manual page the %u format specification is of type int. This means that printf reads the value as int, which even on 64-bit platforms are still 32 bits.

Add a size specification of either ll or I64, so the format will be %I64u.

Answer 3

It should be -1LL because -1 is stored as a 32bit int