hm....
which gcc's version/build?
~d
On Monday 02 February 2004 18:10, G Halkes wrote:
> Hi,
>
> What I ment was, if you use the following code:
>
> struct unaligned {
> char first;
> int second;
> } __attribute__((packed));
>
> union test {
> struct unaligned data;
> char cdata[3];
> };
>
> void function() {
> union test test;
> int data, tmp;
> test.cdata[0] = 0x1;
> test.cdata[1] = 0x2;
> test.cdata[2] = 0x3;
>
> data = test.data.second;
> memcpy(&tmp, &test.data.second, 2);
> }
>
> data will contain the value 0x201, and not the value
> 0x302 as it should. The same goes for tmp (try it. I
> did and got the wrong results). The reason the memcpy
> goes wrong is because it is replaced by a single word
> move instead of two byte moves.
>
> When I built gcc with the STRICT_ALIGNMENT macro set
> to 1, I got the value 0x302 which is correct.
>
> G. Halkes
>
> PS: I used a printf routine to output the results to a
> serial port to see the result. I removed it from the
> code because it isn't standard. Just in case your
> wondering.
>
> >for built-ins gcc takes into account alignement.
> > So, if gcc knows, that something aligned by 2 it
>
> uses >world operations,
>
> > otherwise byte ones.
> >
> > Packed does not mean the structure misaligned.
> > It depends on data types of the structure.
> > For example:
> >
> > struct {
> > char a;
> > int b;
> > };
> >
> > isn't aligned by world boundary.
> >
> > struct { int a,b}
> > aligned.
> >
> > Even more -- if the structure aligned and its size
>
> is >not even, gcc will use
>
> > world operations for thhis structure size()-1 and
> >byte operation for a last
> > byte of it.
>
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