问题描述
大家好，
我得到了一个非常基本的问题，我在这里：
假设我有11001 11010位是真的10位。现在我想
地址，所以如果它是零，我会加一个，如果是一个
那么我将加零。人们说这是曼彻斯特的编码。
请注意，左手边只需接受一下每一次
操作。
如何做到这一点，谢谢。
Hi every one,
I got very basic question, here i go:
Say i have 11001 11010 bits which are infact 10 bits. Now i want to
address every bit so if it is zero i would add one and if it is one
then i would be adding zero. Folks say it manchester coding.
Please note that left hand side just accept a single bit for every
operation.
how to do this , thanks.
推荐答案
我会帮忙，而不是为你工作。
unsigned long ManCode（unsigned long const x ）
{
unsigned long result = x;
for（unsigned i = 0; i< 10; ++ i）
{
if（！GetBit（x，i））++ result;
}
返回结果;
}
-Tomás
I''ll give a hand, rather than do the work for you.
unsigned long ManCode(unsigned long const x)
{
unsigned long result = x;
for(unsigned i = 0; i < 10; ++i)
{
if ( ! GetBit(x,i) ) ++result;
}
return result;
}
-Tomás
我不太确定你要做什么。你是想尝试1'的
赞美比特流吗？
11001 11010变为00110 00101？简单或与1.
0 | 1 = 1
1 | 1 = 0
或者你想做一些与众不同的事情吗？
I''m not quite sure what you''re trying to do. Are you trying to 1''s
compliment the bit stream?
11001 11010 becomes 00110 00101 ? Simply or the bits with 1.
0 | 1 = 1
1 | 1 = 0
Or are you trying to do something different?
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