size of an address when compiling with -m64

[Andrew Haley]

John Fine wrote:
> David Livshin wrote:
>>
>>    double a[DIM][DIM];
>>
>>
>>        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 )
> In other words, the compiler has assumed that the global data object "a"
> will be linked in the first 4GB of the address space.
> 
> If it turns out not to be, then you will get an error at link time.
> 
> I have read (but not really understood) various things about "memory
> model" restrictions in x86_64 that imply rules.  I expect (but I'm not
> sure) one such rule is that certain globals in the main image are
> restricted to the first 4GB.

Yes.

http://gcc.gnu.org/onlinedocs/gcc-4.3.2/gcc/i386-and-x86_002d64-Options.html#i386-and-x86_002d64-Options

See -mcmodel=

Andrew.