多少个值可以被重新使用n比特psented $ P $？ [英] How many values can be represented with n bits?

问题描述

例如，如果 N = 9 ，那么有多少不同的值可以重新$ P $ 9二进制数字psented（位）？

For example, if n=9, then how many different values can be represented in 9 binary digits (bits)?

我的想法是，那些每9位的，如果我设置为1，我会次数最多的可能是那些9位可以重新present。因此，最高值为 111111111 等于十进制511 。我的结论是，因此，二进制的9位可以重新present 511不同的值。

My thinking is that if I set each of those 9 bits to 1, I will make the highest number possible that those 9 digits are able to represent. Therefore, the highest value is 1 1111 1111 which equals 511 in decimal. I conclude that, therefore, 9 digits of binary can represent 511 different values.

是我的思维过程是否正确？如果没有，可能有人善意解释我缺少的是什么？我怎么可以把它推广到 N 位？

Is my thought process correct? If not, could someone kindly explain what I'm missing? How can I generalize it to n bits?

推荐答案

2 ⁹ = 512的值，因为这是你可以有多少的0和1的组合都有。

2⁹ = 512 values, because that's how many combinations of zeroes and ones you can have.

什么这些值重新present但是取决于你所使用的系统上。如果它是一个无符号整数，你将有：

What those values represent however will depend on the system you are using. If it's an unsigned integer, you will have:

000000000 = 0 (min)
000000001 = 1
...
111111110 = 510
111111111 = 511 (max)

在二补时，其通常用于重新二进制present整数，你会有：

In two's complement, which is commonly used to represent integers in binary, you'll have:

000000000 = 0
000000001 = 1
...
011111110 = 254
011111111 = 255 (max)
100000000 = -256 (min) <- yay integer overflow
100000001 = -255
...
111111110 = -2
111111111 = -1

在一般情况下，用的 K 的位可以重新present 2 ^K 值。它们的范围将取决于你所使用的系统上：

In general, with k bits you can represent 2^k values. Their range will depend on the system you are using:

无符号：0〜2 ^K -1结果
  签名：-2 ^K-1 2 ^K-1 1

Unsigned: 0 to 2^k-1
Signed: -2^k-1 to 2^k-1-1

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