----- Original Message -----
From: David Neary <dneary at informix.com>
To: RenWei <renwei at huawei.com>
Cc: <ilug at linux.ie>
Sent: Wednesday, November 08, 2000 7:49 PM
Subject: Re: [ILUG] gcc compile options
> RenWei wrote:
> > hi ,
> > i want to know how "align" works in gcc .
> > i use this in my program, run in a X86 box, RedHat 6.2 full installed:
> > int a __attribute__ ((aligned(8)));
> > printf("size of a : %d \n", sizeof(a));
> > i want it's size be 8 bytes, by i get 4.
> > anyone can tell me how to use align in detail?
>> The size of an int will always be 4 bytes. However, if you were to
> define a struct with one integer element, I believe that the size of the
> struct would be 8. align affects the amount of padding space set aside
> in memory after the end of that variable, so if a is at address 0x0, the
> next variable will be stored (at earliest) in 0x8. But the int is still
> of size 4. If you want ints that are 8 bytes long, you should probably
> use long.
>> As an aside, is an int always the same size as a memory address (ie. 4
> bytes on 32 bit machines, 8 bytes on 64 bit), or is it always 4 bytes?
>> > thanks
> > renwei.
i tried this , it worked.
char a __attribute__((aligned(4));
short b __attribute__((aligned(4));
char c __attribute__((aligned(4));
this case worked. the size of mt is 12 bytes.
but this one:
char a ;
short b ;
char c ;
} mt __attribute__((aligned(4));
does't work. size of me is 6 bytes.
2. i have many struct in some c file to do this ,
Can i use some set global ? once set , all the struct
in c file aligned in 4 bytes.
is there any way to get it?
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