How do you determine 32 or 64 bit architecture of Windows using Java?

Question

How do you determine 32 or 64 bit architecture of Windows using Java?

Answer 1

Maybe it 's not the best way, but it works.

All I do is get the "Enviroment Variable" which windows has configured for Program Files x86 folder. I mean Windows x64 have the folder (Program Files x86) and the x86 does not. Because a user can change the Program Files path in Enviroment Variables, or he/she may make a directory "Program Files (x86)" in C:\, I will not use the detection of the folder but the "Enviroment Path" of "Program Files (x86)" with the variable in windows registry.

public class getSystemInfo {

    static void suckOsInfo(){

    // Get OS Name (Like: Windows 7, Windows 8, Windows XP, etc.)
    String osVersion = System.getProperty("os.name");
    System.out.print(osVersion);

    String pFilesX86 = System.getenv("ProgramFiles(X86)");
    if (pFilesX86 !=(null)){
        // Put here the code to execute when Windows x64 are Detected
    System.out.println(" 64bit");
    }
    else{
        // Put here the code to execute when Windows x32 are Detected
    System.out.println(" 32bit");
    }

    System.out.println("Now getSystemInfo class will EXIT");
    System.exit(0);

    }

}