On Saturday, 25 March 2017 at 22:54:30 UTC, zabruk70 wrote:
But for clearness...I was thinked, that align not changes SIZE, but changes LOCATION.I was thinked, that "align(X) union Union1" just force compiler to place Union1 on boundaries of X bytes...
In order for all Union1 instances in an array to be aligned on boundaries of X bytes, the size of Union1 needs to be padded to a multiple of X in such a case.
