How to distinguish WIndows Architecture

Hi all,
Is there any CMake variable for distinguish, on Windows, if we are building for 32 bits Architecture or 64 bits Architecture?
Thanks in advance for any help

Have a look at CMAKE_SIZEOF_VOID_P variable.

    message(STATUS "32 bit")

    message(STATUS "64 bit")
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Ok, but if I am on a 64 bits Host machine, but I build with -A 32 bits flag, the sizeof of the word is 4 bytes or 8 bytes? I think it will be the latter…

The variable CMAKE_SIZE_OF_VOID_P describes your build environment, not the host on which you are running… So it will be 4 bytes.

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On Windows you basically have 4 different compilers (for x86/AMD64):

# C:\ProgramData\Microsoft\Windows\Start Menu\Programs\Visual Studio 2022\Visual Studio Tools\VC

# Host 64 -> Target 64
x64 Native Tools Command Prompt for VS 2022
# Host 64 -> Target 32
x64_x86 Cross Tools Command Prompt for VS 2022

# Host 32 -> Target 64
x86_x64 Cross Tools Command Prompt for VS 2022
# Host 32 -> Target 32
x86 Native Tools Command Prompt for VS 2022

I never recommend using the last 2 for larger applications since big projects tend to use more than 4 GB of RAM.

Anyway CMake lets you pick both your host tools and your target platform. They are different things.

From the CMake docs:

The CMAKE_GENERATOR_PLATFORM variable may be set, perhaps via the cmake(1) -A option, to specify a target platform name (architecture). For example:

- cmake -G "Visual Studio 17 2022" -A Win32
- cmake -G "Visual Studio 17 2022" -A x64
- cmake -G "Visual Studio 17 2022" -A ARM
- cmake -G "Visual Studio 17 2022" -A ARM64

If you want to choose different host tools here are the docs. However, VS2019 and newer use 64-bit tools by default.

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