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Home/ Questions/Q 8119085
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Editorial Team
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Editorial Team
Asked: June 6, 20262026-06-06T04:41:47+00:00 2026-06-06T04:41:47+00:00

C++ code (Visual studio started with devenv /useenv (x64) and isWOW64 is false) DWORD64

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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.

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  1. Editorial Team
    Editorial Team
    2026-06-06T04:41:49+00:00Added an answer on June 6, 2026 at 4:41 am

    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.

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