"<c Language adapter and Implementation", "code of a doubt In Chapter 13 "vectors" section of the code are : # Nbytes (practically) ((((practically) + 8 - 1) & (~ (8 - 1))) /8) According to the book means that the Acer used to calculate [len/8], the question is : 1. I have seen this code to write and direct len/8 find no difference between Asia because : (Practically + 8 - 1) & (~ (8 - 1)), is len+7 front, behind and on 7 premium is not that of seven away? 280-1 = 7, then why should be rewritten 8-1? 3. I carefully thought for a moment, think that this issue should be the exception and integer calculation, in the end what? "<c Language adapter and Implementation", "code of a doubt Of course there is a distinction. This procedure, if len<8, is the one and len/8 integer, if the len<8 is 0. "<c Language adapter and Implementation", "code of a doubt "<c Language adapter and Implementation", "code of a doubt In fact, nbytes (n) = n ／ 8 +1 And If len<8, are given 1; No contradiction! I saw the wording to read, but I have no problem 8-1 and 7 clear; Repeat tomorrow ask the Supreme find! "<c Language adapter and Implementation", "code of a doubt Give you another example : N = 9; N/8==1; (N+7) /8 ==2; This is to take the whole, and not lose spaces (N+7) &&! (8-1) and went to the same role "<c Language adapter and Implementation", "code of a doubt Right there! "<c Language adapter and Implementation", "code of a doubt (Len+8-1) & (~ (8-1)) = (len+7) /8, meaning two months is practically required 229 spaces occupied by the byte count. After a written translation of the former than the latter wording of computational efficiency. Written (8-1), which was to improve the readability and easy replacement. For example, eight replaced 16, it becomes practically two months short of the 229 spaces occupied by the few. "<c Language adapter and Implementation", "code of a doubt Agree bierdaci "<c Language adapter and Implementation", "code of a doubt --> So that it makes sense [ Closed window ]