parsing precedence of long integer literal "L"

[Andrew Chi <achi@xxxxxxx>]

I'm confused about the parsing of the "L" which should denote a long 
integer literal.  I'm running GCC 4.5.3 on a 32-bit machine, i.e. long 
long = 64 bits and long = 32 bits.  Am I misunderstanding the precedence 
of "L" or how it's parsed/converted?

====================================================================
Sample code:

#include <stdio.h>

int main(void)
{
  long long x = 10;
  long long y = (0xffffffffL);
  long long z = (long)(0xffffffffL);

  printf("long long x == %lld\n", x);
  printf("long long y == %lld\n", y);
  printf("long long z == %lld\n", z);

  printf("0xffffffffL == %ld\n", 0xffffffffL);

  if (x > (long)(0xffffffffL))
    printf("x > (long)(0xffffffffL)\n");
  else
    printf("x <= (long)(0xffffffffL)\n");

  if (x > (0xffffffffL))
    printf("x > (0xffffffffL)\n");
  else
    printf("x <= (0xffffffffL)\n");
  return 0;
}

====================================================================
Output:

long long x == 10
long long y == 4294967295
long long z == -1
0xffffffffL == -1
x > (long)(0xffffffffL)
x <= (0xffffffffL)

====================================================================
System version (checked on both Cygwin and Debian Linux):

$ gcc --version; uname -a
gcc (GCC) 4.5.3
Copyright (C) 2010 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.

CYGWIN_NT-6.1 FOOBAR 1.7.17(0.262/5/3) 2012-10-19 14:39 i686 Cygwin