C# Directive to indicate 32-bit or 64-bit build

Question

Is there some sort of C# directive to use when using a development machine (32-bit or 64-bit) that says something to the effect of:

if (32-bit Vista)
    // set a         


        

Answer 1

What I use in my C# code is IntPtr.Size, it equals 4 on 32bit and 8 on 64bit:

string framework = (IntPtr.Size == 8) ? "Framework64" : "Framework";

Answer 2

You can use Predefined Macros to set the properties on compilation

    #if (_WIN64) 
  const bool IS_64 = true;
#else
  const bool IS_64 = false;
#endif

Answer 3

You can use a #if directive and set the value as a compiler switch (or in the project settings):

 #if VISTA64
     ...
 #else
     ...
 #endif

and compile with:

 csc /d:VISTA64 file1.cs 

when compiling a 64 bit build.

Answer 4

There are two conditions to be aware of with 64-bit. First is the OS 64-bit, and second is the application running in 64-bit. If you're only concerned about the application itself you can use the following:

if( IntPtr.Size == 8 )
   // Do 64-bit stuff
else
   // Do 32-bit

At runtime, the JIT compiler can optimize away the false conditional because the IntPtr.Size property is constant.

Incidentally, to check if the OS is 64-bit we use the following

if( Environment.GetEnvironmentVariable( "PROCESSOR_ARCHITEW6432" ) != null )
    // OS is 64-bit;
else
    // OS is 32-bit

Answer 5

There is nothing built in that will do this for you. You could always #define your own symbol and use that for conditional compilation if you wish.

Answer 6

Can you do it at runtime?

if (IntPtr.Size == 4)
  // 32 bit
else if (IntPtr.Size == 8)
  // 64 bit