一个字节数组的高效二进制补码 [英] efficient two's complement of a byte array

问题描述

计算可变长度字节数组的两个补码

的最有效方法是什么？

感谢您的时间。

What would be the most efficient way to calculate the two''s complement
of a variable length byte array?
Thanks for your time.

推荐答案

da ***** **** @gmail.com 写道：

什么是计算两个补码的最有效方法

的可变长度字节数组？

What would be the most efficient way to calculate the two''s complement
of a variable length byte array?

我会假设一个for循环否定就足够了

用于大多数目的。


你可以尝试否定int或者长期使用不安全的代码和

a指针。


Arne

I would assume a for loop negating would be good enough
for most purposes.

You could try negating int or long using unsafe code and
a pointer.

Arne

2月10日下午12:05，Arne Vajh？j< a ... @ vajhoej.dkwrote：

On Feb 10, 12:05 pm, Arne Vajh?j <[e-mail]:

darthgha ... @ gmail.com写道：

[e-mail] wrote:

计算可变长度字节数组的两个补码

的最有效方法是什么？

What would be the most efficient way to calculate the two''s complement
of a variable length byte array?

我会假设一个for循环否定就足够了

用于大多数目的。


你可以尝试否定int或者长期使用不安全的代码和

a指针。


Arne

I would assume a for loop negating would be good enough
for most purposes.

You could try negating int or long using unsafe code and
a pointer.

Arne

我不能否定int或long，因为数字可能超过40亿

位数（十六进制）。 for循环可以解决这个问题，但我希望有人可能知道不同的方式。谢谢你的想法

虽然。

克里斯

I can''t negate int or long because the number could be over 4 billion
digits (in hex). A for loop would do the trick, but I was hoping
someone might know of a different way. Thanks for the thoughts
though.
Chris

2月12日，下午2:54 ，darthgha ... @ gmail.com写道：

On Feb 12, 2:54 pm, [e-mail] wrote:

2月10日下午12:05，Arne Vajh？j< a ... @ vajhoej.dkwrote ：

On Feb 10, 12:05 pm, Arne Vajh?j <[e-mail]:

darthgha ... @ gmail.com写道：

[e-mail] wrote:

什么是最有效的方式计算一个可变长度字节数组的两个补码

？

What would be the most efficient way to calculate the two''s complement
of a variable length byte array?

我会假设一个for循环否定就足够了

用于大多数目的。

I would assume a for loop negating would be good enough
for most purposes.

您可以尝试使用不安全的代码来否定int或者长期使用

指针。

You could try negating int or long using unsafe code and
a pointer.

Arne

Arne

我不能否定int或long因为数字可以超过40亿

位数（十六进制）。 for循环可以解决这个问题，但我希望有人可能知道不同的方式。谢谢你的想法

虽然。

I can''t negate int or long because the number could be over 4 billion
digits (in hex). A for loop would do the trick, but I was hoping
someone might know of a different way. Thanks for the thoughts
though.

Arne建议的是一个古老的C技巧：循环遍历数组

一次一个长（64位），否定每个很长一段时间，而不是这样做。
a字节一次。但是，正如他所指出的那样，你必须使用

不安全代码才能做到这一点。

What Arne was suggesting was an old C trick: looping over the array
one long (64 bits) at a time, negating each long, rather than doing it
a byte at a time. However, as he pointed out, you would have to use
unsafe code to do this.

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