How infinite numbers are represented under limitation of finite bits? In this blog, we focus on representations of integers and floating point numbers.
Integers
Above is the philosophy of the binary representation. Now we look further into specific representation of integers. C/C++ standards provide two different ways: one for nonnegative numbers (unsigned encodings), the other for negative, zero, and positive numbers (signed encodings).
Given a 0-1 sequence $B = x_{w-1}x_{w-2}…x_{1}x_{0}$, its corresponding value is $\sum_{i=0}^{w-1} x_i2^i$ in unsigned encodings. As a result, it forms a bijection. Every number between $0$ and $2^w − 1$ has a unique encoding as a $w$-bit value, while negative numbers are not included.
Therefore in signed encodings, we interpret the value as $-x_{w-1}2^{w-1} + \sum_{i=0}^{w-2} x_i2^i$. The different lies and only lies on the most significant bit (MSB). It is still a bijection, but the range now becomes $[-2^{w-1}, 2^{w-1} - 1]$. For nonnegative number x, its representation is the same for these two encodings with MSB as 0, while for negative number x, its representation will be interpreted as $2^w + x$, which is the complement of it with respect to $2^w$. It is exactly why we also call this encoding Twos’ complement encodings.
It comes to us naturaly that whether there are other encodings to represent signed numbers. The answer is of course yes. Remember, we can interpret a sequence in any way. We can enfranchise one bit any meaning we want. For example, in so-called Ones’ comlement encodings, MSB’s weight is $-(2^{w-1} - 1)$ and the value is calculated as $-x_{w-1}(2^{w-1}-1) + \sum_{i=0}^{w-2} x_i2^i$. It has the range of $[-2^{w-1} - 1, 2^{w-1} - 1]$. Is it still a bijection? No, because zero can be represented as $000…000$ or $111…111$ in this way.
Floating point Numbers
Floating point numbers are more difficult to represent.
You can learn more from CMU course 15-213 or its textbook CS:APP.
