给定N位，可以用二进制表示多少个整数? [英] Given N bits, how many integers can be represented in binary?

问题描述

假设您有14位.您如何确定从这14位中可以用二进制表示多少个整数?

Suppose you have 14 bits. How do you determine how many integers can be represented in binary from those 14 bits?

仅仅是2 ^ n吗?所以2 ^ 14 = 16384?

Is it simply just 2^n? So 2^14 = 16384?

请注意问题的这一部分:二进制中可以表示多少个整数...".那就是我困惑的地方，否则这个问题看起来很简单.如果问题只是问从14位可以表示多少个不同的值或数字，那么肯定会是2 ^ n.

Please note this part of the question: "how many INTEGERS can be represented in BINARY...". That's where my confusion lies in what otherwise seems like a fairly straightforward question. If the question was just asking how many different values or numbers can be represented from 14 bits, than yes, I'm certain it's just 2^n.

推荐答案

答案取决于您需要带符号整数还是无符号整数.

The answer depends on whether you need signed or unsigned integers.

如果需要无符号整数，则可以使用 2 ^ n 表示从0到2 ^ n的整数.例如n = 2；2 ^ 2 = 4可以表示0到4之间的整数(不包括0到3).因此，使用n位，您可以表示最大无符号整数值为 2 ^ n-1 ，但总数为 2 ^ n 包括0的整数.

If you need unsigned integers then using 2^n you can represent integers from 0 to 2^n exclusive. e.g. n=2; 2^2=4 you can represent the integers from 0 to 4 exclusive (0 to 3 inclusive). Therefore with n bits, you can represent a maximum unsigned integer value of 2^n - 1, but a total count of 2^n different integers including 0.

如果需要带符号的整数，则一半的值是负数，一半的值是正数，并且1位用于指示整数是正数还是负数.然后，您可以使用 2 ^ n/2 进行计算.例如n = 2；2 ^ 2/2 = 2您可以表示-2到2的整数(不包括-2到+1).0被认为是正值，因此您将获得2个负值(-2，-1)和2个正值(0和+1).因此，使用n位，您可以表示(-)2 ^ n/2 和(+)2 ^ n/n-1 ，但您仍然拥有与无符号整数相同的总数 2 ^ n 个不同的整数.

If you need signed integers, then half of the values are negative and half of the values are positive and 1 bit is used to indicate whether the integer is positive or negative. You then calculate using using 2^n/2. e.g. n=2; 2^2/2=2 you can represent the integers from -2 to 2 exclusive (-2 to +1 inclusive). 0 is considered postive, so you get 2 negative values (-2, -1) and 2 positive values (0 and +1). Therefore with n bits, you can represent signed integers values between (-) 2^n/2 and (+) 2^n/n - 1, but you still have a total count of 2^n different integers as you did with the unsigned integers.

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