size of an address when compiling with -m64

[David Livshin <david.livshin@xxxxxxxxxxx>]

Hi,

Using gcc 4.4 cross compiler ( running on IA-32 but generating code for 
x86-64 ):

   gcc  -O3 -fomit-frame-pointer -ffast-math  -mfpmath=sse -msse2 
-msse4 -mmmx   -S -m64 -o x.s x.c

for x.c being:

   #define DIM 1000
   double a[DIM][DIM];

   double dot()
   {
       double sum;
       int i, j;

       sum = 0.;
       for ( i = 0; i < DIM; i++ )
           {
               for ( j = 0; j < DIM; j++ )
                   {
                       sum += a[i][j];
                   }
           }

       return sum;

   }


generates the following assembly output x.s ( only part is shown ):

   .type dot, @function
   dot:
   .LFB0:
       xorpd %xmm0, %xmm0
       movl $a, %eax

          .
          .


The register 'eax' ( 'rax' actualy ) is used later in the code as the 
base for an address.
Questions:

  1. Why 'movl'  and not 'movq' is used? ( It seems that compilers
     assumes that the size of the address of the symbol 'a' is 32 bits
     only )
  2. How to force the compiler to generate 'movq'?


Thanks in advance.

--
David Livshin

http://www.dalsoft.com