Question

I’m trying to create a basic function to swap the endianness of a short int but it’s throwing an error:

#ifndef __ENDIAN__
#define __ENDIAN__

#define Swap16(value) \
    ((((unsigned short)((value) & 0x00FF)) << 8) | \
    (((unsigned short)((value) & 0xFF00)) >> 8))

#define Swap32(value) \
    ((((unsigned)((value) & 0x000000FF)) << 24) | \
    (((unsigned)((value) & 0x0000FF00)) << 8) | \
    (((unsigned)((value) & 0x00FF0000)) >> 8) | \
    (((unsigned)((value) & 0xFF000000)) >> 24))

void __inline SwapEndian(short* value) //ERROR HERE
{
    *value = Swap16(value);
}

#endif

I intend to use the code like:

short val = 0x1234;
SwapEndian(&val);
//val now contains 0x3412

I’m using VS2008 and the exact error is:

C2296: '&' : illegal, left operand has type 'short *'

What can I do to fix this error?

Answer 1

If you look at how the macro is expanded it’s pretty clear what’s wrong.

&val is a pointer of type short*. But you macro does arithmetic directly off of it’s parameter. So you’re trying to perform integer arithmetic on a pointer.

So this gets expanded to:

((((unsigned short)((&val) & 0x00FF)) << 8)
...

&val is type short*, while the macros expects an integer.

In the function you’ll need to dereference it into an integer before passing it in to the macro.