What does "#elif with no expression" mean?

Question

I am trying to compile a program (that I did not write) and I get the following error:

C read.c ...
In file included from read.c:6:0:
def.h:6:6: error: #elif with no expression
make: *** [read.o] Error 1

File def.h looks like this:

#ifndef TRACE_DEF
#define TRACE_DEF

#ifndef L
  #define L 152064 /* (352 * 288 * 1.5) */
#elif
  #error "L defined elsewhere"
#endif

#ifndef MIN
  #define MIN(a, b) ((a) < (b) ? (a) : (b))
#endif
#ifndef MAX
  #define MAX(a, b) ((a) > (b) ? (a) : (b))
#endif

Line 6 is the line just before #error "L defined elsewhere".

Compiler is:

$ gcc --version
gcc-4.6.real (Ubuntu/Linaro 4.6.3-1ubuntu5) 4.6.3
Copyright (C) 2011 Free Software Foundation, Inc.
This is free software; see the source for copying conditions.  There is NO
warranty; not even for MERCHANTABILITY or FITNESS FOR A PARTICULAR PURPOSE.

Any ideas how to fix it?

Answer 1

Because #elif expects an expression, just like #if. You want to use #else. Otherwise you have to give the expression:

#ifndef L
  #define L 152064 /* (352 * 288 * 1.5) */
#elif defined(L)
  #error "L defined elsewhere"
#endif

(equivalent)

#ifndef L
  #define L 152064 /* (352 * 288 * 1.5) */
#else
  #error "L defined elsewhere"
#endif