On Feb 9 2007 15:29, linux-os (Dick Johnson) wrote:
>>
>> I was referring to "absolute memory", not the offset magic that assembler
>> allows. After all, (reg+relativeOffset) will yield an absolute address.
>> What I was out at: for machines that have more than 2 GB of memory, you
>> don't call the address that is given by 0x80000000U actually "byte
>> -2147483648", but "byte 2147483648".
>
>Don't make any large bets on that!
>
>char foo()
>{
> volatile char *p = (char *)0x80000000;
> return *p;
>}
>Optimized....
> .file "zzz.c"
> .text
> .p2align 2,,3
>.globl foo
> .type foo, @function
>foo:
> pushl %ebp
> movb -2147483648, %al
> movl %esp, %ebp
> movsbl %al,%eax
> leave
> ret
> .size foo, .-foo
> .section .note.GNU-stack,"",@progbits
> .ident "GCC: (GNU) 3.3.3 20040412 (Red Hat Linux 3.3.3-7)"
00000000 <foo>:
0: 55 push %ebp
1: 0f b6 05 00 00 00 80 movzbl 0x80000000,%eax
8: 89 e5 mov %esp,%ebp
a: 5d pop %ebp
b: 0f be c0 movsbl %al,%eax
e: c3 ret
You do know that there is a bijection between the set of signed [32bit]
integers and unsigned [32bit] integers, don't you?
For the CPU, it's just bits. Being signed or unsigned is not important
when just accessing memory. It will, when a comparison is involved, but
that was not the point here. void* comparisons are unsigned. Period.
Because a compiler doing signed comparisons will "map" the memory [from
2 GB to 4 GB] as part of the signed comparison before the memory [from 0
GB to 2 GB], which collides with - let's call it - "the world view".
Jan
--
ft: http://freshmeat.net/p/chaostables/
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