Because it is a compile-time operator that, in order to calculate the size of an object, requires type information that is only available at compile-time. This doesn't hold for C++.
i have read this statement roght now when implementing one library function. why it doesn't hold for C++?
secondly, i am finding it get one thing : "avavilable at compile-time only" ? can anyone explain it to me ? thanks.
Edited 4 Years Ago by nitin1