Re: How to make local variable's address is 8 byte alignment?

David Mathog wrote:
Pengjun Jia wrote:
HP-UX 11.23, HP aC++ B3910B A.03.63

Here is a samples.

bjhp1 /nfs/users/pjia>cat tt.c
#include <stdlib.h>

int main()
{
int i =10 ;
char p[1024] ="" ;

printf("%p, %p\n", &i, p) ;
return 0 ;
}

Well, you could always allocate an extra 8 bytes for the array p
and then use a pointer to the first position in that array which is on an 8 byte alignment instead of p itself.

I don't see the point though for this example. With integers and such
it will likely make a difference if the variable is not properly aligned, but the compiler will generally take care of that for you.

If the OP wanted to use a general area of bytes as a place he could take a pointer into and manipulate as (for example) long or double, he may
need this sort of guarantee. Perhaps he'll tell us what he's trying to
do...
.

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