Re: size of an address when compiling with -m64

[John Fine <johnsfine at verizon dot net>]

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.
>   2. How to force the compiler to generate 'movq'?
I know you can use -fPIC when compiling code for .so files to turn off 
some of the compile time assumptions about link time operations.  I 
assume you're not compiling a .so, so -fPIC is probably drastic (if not 
flat out incorrect) for your purposes.

Do you really think that 8MB data object needs to be linkable outside 
the first 4GB?