On Sat Oct 01, 2005 at 18:23:40 +1000, O Plameras wrote:
>Matthew Hannigan wrote:
>
>>On Fri, Sep 30, 2005 at 10:09:40AM +1000, O Plameras wrote:
>>
>>
>>>.... and as C is closely bound
>>>to hardware architecture you must have said something about these data
>>>
>>>
>>
>>Actually, C is not necessarily that closely bound to hardware architecture.
>>
>>
>>
>The the following C program illustrates the relationship of C and
>hardware architecture on
>16-bit and 32-bit (Two different architectures). We cannot use 32-bit
>and 64-bit (AMD) to
>illustrate because the int size are the same in both. This is an
>aberration for AMD CPUs, I
>think.
>
>Many documentations says that a 64-bit should really have int size=8
>instead of 4
>
>(see http://www.osdata.com/topic/language/asm/datarep.htm
>-DEC VAX *16* *bit* [2 *byte*] *word*; 32 bit [4 *byte*] longword; 64
>bit [8 *byte*]quadword;
>132 bit [16 *byte*] octaword; data may be unaligned at a speed penalty *...*
>)
>
>There are other documentations similar to the above assertions on the net.
It depends on the architecture ABI. As others have shown AMD64, Itanium, (and
I'm
pretty sure power5) architectures use 32-bit ints. Otehr 64-bit architectures
may choose differently.
>But check the C-codes.
>
>#include <stdio.h>
>struct verify {
> char initials[2];
> int birthdate;
>};
>int main(void)
>{
> struct verify holes;
> printf ("%d\n", sizeof(holes.initials[0]));
> printf ("%d\n", sizeof(holes.initials));
> printf ("%d\n", sizeof(holes.birthdate));
> printf ("%d\n", sizeof(holes));
> return 0;
>}
>Given that the word-byte
>In 16-bit computer = 2 bytes word the output is,
>1
>2
>2
>4
>in 32-bit computer = 4 bytes word the output is,
>1
>2
>4
>8
>
>Two things are different due to computer architecture.
>1. The same struct have different memory sizes allocated.
Yep.
>2. The struct has no hole in 16-bit and has 2-byte hole in 32-bit
>because C-compiler always
> align int data types beginning at the next word byte.
I don't think the C standard specifies how a struct should be layed out
in memory.
Matthew said is wasn't `necessarily' bound to the hardware. In practise you
are going to see different data type sizes on different bits of hardware.
But just because the hardware can store an 64-bit words, doesn't mean that
the ABI is going to specify an int to be 64-bit.
Benno
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