A＆QUOT;动态位域＆QUOT;用C

本文介绍了A＆QUOT;动态位域＆QUOT;用C的处理方法，对大家解决问题具有一定的参考价值，需要的朋友们下面随着小编来一起学习吧！ 问题描述

在这个问题中，假设所有的整数都是无符号为简单起见。

In this question, assume all integers are unsigned for simplicity.

假如我还想写两个功能，包并解压，这让你收拾宽度较小的整数成，也就是说，一个64位整数。然而，整数的位置和宽度在运行时给出，所以我不能用C位域。

Suppose I would like to write 2 functions, pack and unpack, which let you pack integers of smaller width into, say, a 64-bit integer. However, the location and width of the integers is given at runtime, so I can't use C bitfields.

最快的是用一个例子来解释。为简单起见，我将与8位整数说明：

Quickest is to explain with an example. For simplicity, I'll illustrate with 8-bit integers:

* * bit # 8 7 6 5 4 3 2 1 myint 0 1 1 0 0 0 1 1

假设我想解压在位置5，宽度为2。这些整数都标有星号的两位。该操作的结果，应0B01。同样的，如果我解压在位置2，宽6，我会得到0b100011。

Suppose I want to "unpack" at location 5, an integer of width 2. These are the two bits marked with an asterisk. The result of that operation should be 0b01. Similarly, If I unpack at location 2, of width 6, I would get 0b100011.

我可以用bitshift左其次是bitshift右随便写的解压功能。

I can write the unpack function easily with a bitshift-left followed by a bitshift right.

但我想不出写一个相当于包的功能，它会做相反的一条明路。

But I can't think of a clear way to write an equivalent "pack" function, which will do the opposite.

说给出的整数0b11，在位置5它包装成敏（上面）和宽度2将产生

Say given an integer 0b11, packing it into myint (from above) at location 5 and width 2 would yield

* * bit # 8 7 6 5 4 3 2 1 myint 0 1 1 1 0 0 1 1

我想出了最佳涉及到很多concatinating位串与OR的，＆LT;＆LT;和>>。之前，我实现和测试它，也许有人认为一个聪明的快速解决方案？

Best I came up with involves a lot of concatinating bit-strings with OR, << and >>. Before I implement and test it, maybe somebody sees a clever quick solution?

推荐答案

关闭我的头顶，未经考验的。

Off the top of my head, untested.

int pack(int oldPackedInteger, int bitOffset, int bitCount, int value) { int mask = (1 << bitCount) -1; mask <<= bitOffset; oldPackedInteger &= ~mask; oldPackedInteger |= value << bitOffset; return oldPackedInteger; }

在您的例子：

int value = 0x63; value = pack(value, 4, 2, 0x3);

要在写入一个值34（可带两位）的偏移时，0x63是当前值。

To write the value "3" at an offset of 4 (with two bits available) when 0x63 is the current value.