Re: A portable way of pointer alignment
"Alf P. Steinbach" <alfps@start.no> wrote in message
news:hm74pd$rk7$1@news.eternal-september.org...
* Chris M. Thomasson:
"Chris M. Thomasson" <no@spam.invalid> wrote in message
news:HlEhn.17$NH1.0@newsfe14.iad...
"Alf P. Steinbach" <alfps@start.no> wrote in message
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[...]
[...]
I don't think `offsetof()' works with POD.
YIKES! I meant to say:
I don't think `offsetof()' works with non-POD types.
;^o
Formally it doesn't, in practice it may and probably will, depending on
the compiler.
I am getting some warnings, however the end result seems perfectly fine with
me:
_________________________________________________________________
#include <cstddef>
#include <iostream>
template< typename T >
inline std::size_t alignmentOf()
{
struct X
{
char bah;
T t;
};
return offsetof( X, t );
}
struct object_base
{
char m_1[2];
double m_d[6];
long m_2[2];
short m_3[14];
virtual void foo() = 0;
};
struct object : public object_base
{
char m_1x[2];
void foo()
{
}
};
int
main()
{
std::cout << "alignmentOf<object> == "
<< alignmentOf<object>()
<< std::endl;
return 0;
}
_________________________________________________________________
Here are the warnings from gcc:
_________________________________________________________________
In function `size_t alignmentOf() [with T = object]':
instantiated from here
[Warning] invalid access to non-static data member `alignmentOf() [with T =
object]::X::t' of NULL object
[Warning] (perhaps the `offsetof' macro was used incorrectly)
_________________________________________________________________
Still, I am a bit weary of using `offsetof()' on a non-POD type. Perhaps I
am just way to paranoid!
;^D
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