> > > char buf[8]; > > > void *v = &buf[1]; > > > unsigned int *p = (unsigned int *)v; > > > > This does not (reliably) do what you expect. The compiler need not align > > buf. > > Printing the value of p should clarify this. > > And, as we can see above, the "simple" accesses are left to the hardware > to fix up. However, if the misaligned access is performed using a > 64-bit value pointer, then the kernel will trap an exception and the > access will be simulated.
I think you've missed my point. gcc may (though unlikely in this case) choose to place buf at an odd address. In which case p will happen to be properly aligned. I'm not sure where you get "64-bit value pointer" from. *p is only a word sized access, and memcpy is defined in terms of bytes so will only be promoted to wider accesses when the compiler believes it is safe. Paul _______________________________________________ linaro-dev mailing list linaro-dev@lists.linaro.org http://lists.linaro.org/mailman/listinfo/linaro-dev
