n 位可以表示多少个值? [英] How many values can be represented with n bits?

问题描述

例如，如果n=9，那么9个二进制数字(位)可以表示多少个不同的值?

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

我的想法是，如果我将这 9 位中的每一个都设置为 1，那么我将使这 9 位能够表示的最大数字成为可能.因此，最大值是 1 1111 1111，它等于十进制的 511.因此，我得出结论，9位二进制可以代表511个不同的值.

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.

我的思维过程正确吗?如果没有，有人可以解释一下我错过了什么吗?如何将其概括为 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

2⁹ = 512 个值，因为你可以有多少个 0 和 1 的组合.

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

然而，这些值代表什么取决于您使用的系统.如果它是一个无符号整数，您将拥有:

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)

在二的补码中，通常用于表示二进制整数，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

通常，使用 k 位可以表示 2^k 个值.它们的范围取决于您使用的系统:

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 到 2^k-1
签名:-2^k-1 到 2^k-1-1

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

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