Digital-electronics 简明教程

Digital Electronics - Hexadecimal Arithmetic

What is Hexadecimal Arithmetic?

在数字电子中，十六进制数用于以更紧凑形式表示二进制信息，因为一位十六进制数字可以表示一组 4 位二进制数字。因此，十六进制数和对它们执行的算术运算在数字电子领域中扮演着至关重要的角色。

In digital electronics, hexadecimal numbers are used to represent binary information in more compact form, as one hexadecimal digit can represent a group of 4 binary digits. Therefore, hexadecimal numbers and arithmetic operation on them play a vital role in the field of digital electronics.

十六进制算术是一种允许对十六进制或基数为 16 的数字进行加法、减法、乘法和除法等算术运算的数学系统。

Hexadecimal arithmetic is a mathematical system that allows to perform arithmetic operations such as addition, subtraction, multiplication, and division of hexadecimal or base-16 numbers.

在本章中，我们将涵盖以下四个基本的十六进制算术运算：

In this chapter, we will cover the following four basic hexadecimal arithmetic operations −

Hexadecimal Addition
Hexadecimal Subtraction
Hexadecimal Multiplication
Hexadecimal Division

让我们借助示例详细了解每个十六进制算术运算。

Let’s understand each of the hexadecimal arithmetic operations in detail with the help of examples.

Hexadecimal Addition

十六进制加法是对十六进制数执行的基本算术运算之一，用于确定它们的和。基本上，十六进制加法类似于十进制加法。但在十六进制加法中，如果和大于或等于 16，则会产生向较高位进位的情况。

Hexadecimal addition is one of the basic arithmetic operations performed on hexadecimal numbers to determine their sum. Basically, hexadecimal addition is similar to decimal addition. But in hexadecimal addition, a carry is generated to the next higher column if the sum is greater than or equal to 16.

让我们看一些已解决的示例，以更好地理解十六进制加法。

Let us see some solved examples to better understand the hexadecimal addition.

Example 1

对 (5A)16 和 (BF)16 求和。

Add (5A)16 and (BF)16.

Solution

对给定十六进制数的加法如下所示：

The addition of the given hexadecimal numbers is shown below −

(5A)16 + (BF)16 = (119)16

Explanation

从最右列开始对十六进制数字求和：A + F = 10 + 15 = 25 = 16 + 9。此处，16 形成进位到下一列。因此，和为 9，并有 1 进位到下一列。

Start by adding the hexadecimal digits in the rightmost column: A + F = 10 + 15 = 25 = 16 + 9. Here, 16 forms a carry to the next column. Thus, the sum is 9 with a 1 as carry to the next column.

移到下一列，并对数字加上进位求和：5 + B + 1 = 5 + 11 + 1 = 17 = 16 + 1。此处，16 形成进位到下一列。因此，和为 1，并有进位 1。没有剩余的数字，因此进位也将写为和中最高位数字。

Move to the next column and add the digits along with carry: 5 + B + 1 = 5 + 11 + 1 = 17 = 16 + 1. Here, 16 forms a carry to the next column. Thus, the sum is 1 with a carry 1. There are no digits left, hence carry will also be written as leftmost digit in the sum.

因此，5A 和 BF 的十六进制和为 119。

So, the hexadecimal sum of 5A and BF is 119.

Example 2

对 (ABC)16 和 (2A9)16 求和。

Add (ABC)16 and (2A9)16.

Solution

给定数字的十六进制和如下所示：

The hexadecimal sum of given numbers is shown below −

(ABC)16 + (2A9)16 = (D65)16

Explanation

从从最右列开始对数字求和：C + 9 = 12 + 9 = 21 = 16 + 1。此处，16 形成进位。因此，和为 1，并有进位 1。

Start by adding the digits in the rightmost column: C + 9 = 12 + 9 = 21 = 16 + 1. Here, 16 forms a carry. Thus, the sum is 1 with a carry 1.

移到下一列，并对数字加上上一步的进位求和：B + A + 1 = 11 + 10 + 1 = 22 = 16 + 6。因此，和为 6，并有进位 1 到下一列。

Move to the next column and add the digits along with the carry from the previous step: B + A + 1 = 11 + 10 + 1 = 22 = 16 + 6. Thus, the sum is 6 with a carry 1 to the next column.

移到最高位列，并对数字加上上一步的进位求和：A + 2 + 1 = 10 + 2 + 1 = 13。由于和为 13，小于 16，因此不会产生进位。在十六进制数系统中，13 用字母 D 表示。

Move to the leftmost column and add the digits along with the carry from the previous step: A + 2 + 1 = 10 + 2 + 1 = 13. Since, the sum is 13 which is less than 16, hence no carry is generated. In hexadecimal number system, 13 is represented by the letter D.

因此，ABC 和 2A9 的十六进制和 = D65。

Hence, the hexadecimal sum of ABC and 2A9 = D65.

以上内容关于十六进制加法，涉及按列相加给定十六进制数字的数字。在执行十六进制加法时需要记住的最重要一点是，当特定列中的和大于或等于 16（即十六进制数字系统的基数）时，会将进位生成到下一列。

This is all about hexadecimal addition that involves the addition of digits of the given hexadecimal numbers column by column. The most important point to keep in mind while performing hexadecimal addition is that a carry is generated to the next column when the sum in a particular column is greater than or equal to 16, i.e., base of the hexadecimal number system.

Hexadecimal Subtraction

十六进制减法是对十六进制数字执行的一项基本算术运算，用于确定它们之间的差。

Hexadecimal subtraction is a basic arithmetic operation performed on hexadecimal numbers to determine the difference between them.

十六进制减法类似于十进制减法。唯一的区别在于十六进制减法中，当被减数数字小于减数数字时，将从较高的列数位中借用 1，相当于 16。

Hexadecimal subtraction is similar to decimal subtraction. The only difference is that in hexadecimal subtraction, when the minuend digit is smaller than the subtrahend digit, a borrow 1, which is equivalent to 16, is taken from the higher column digit.

让我们借助已解决的示例来理解十六进制减法。

Let us understand the hexadecimal subtraction with the help of solved examples.

Example 1

从 (A57)16 中减去 (125)16。

Subtract (125)16 from (A57)16.

Solution

给定十六进制数字的减法如下所示 −

The subtraction of given hexadecimal numbers is given below −

(A57)16 - (125)16 = (932)16

Explanation

从最右侧列开始减去十六进制数字：7 – 5 = 2。写下结果。

Start subtracting the hexadecimal digits from rightmost column: 7 – 5 = 2. Write down the result.

移至下一列并减去数字：5 – 2 = 3。将数字 3 写为差。

Move to the next column and subtract the digits: 5 – 2 = 3. Write down the digit 3 as difference.

移至最左侧列并减去数字：A – 1 = 10 – 1 = 9。将结果写为差。

Move to the leftmost column and subtract the digits: A – 1 = 10 – 1 = 9. Write down the result as difference.

因此，A57 和 125 的十六进制差为 932。

So, the hexadecimal difference of A57 and 125 is 932.

Example 2

从 (BC5)16 中减去 (1DA)16。

Subtract (1DA)16 from (BC5)16.

Solution

BC5 和 1DA 的十六进制减法如下所示 −

The hexadecimal subtraction of BC5 and 1DA is shown below −

(BC5)16 - (1DA)16 = (9EB)16

Explanation

从最右侧列的数字开始减去：5 – A。由于 5 小于 A（10），因此我们必须从高一级数字中借用。从下一列 © 借用后，数字 5 将变为 5 + 16（因为 16 等于借用 1）= 21。因此，21 – A = 11 (B)。将 B 写为差。

Start by subtracting from the digits in rightmost column: 5 – A. Since 5 is less than A (10), so we have to borrow from the next higher-order digit. After borrowing from the next column ©, the digit 5 will become 5 + 16 (as 16 is equivalent to borrow 1) = 21. Thus, 21 – A = 11 (B). Write down B as the difference.

移至下一列并减去数字：B – D。同样，B 小于 D，因此我们从高位数字 B 中借用。借用后，B 将变为 B + 16 = 27。因此，27 – D = 14 (E)。将数字 E 写为差。

Move to the next column and subtract the digits: B – D. Again, B is smaller than D, so we take a borrow from the higher order digit B. After getting a borrow, B will become B + 16 = 27. Thus, 27 – D = 14 (E). Write down the digit E as difference.

移至最左侧列并减去数字：A – 1 = 9。写下结果。

Move to the leftmost column and subtract the digits: A – 1 = 9. Write down the result.

因此，BC5 和 1DA 的十六进制差等于 9EB。

Hence, the hexadecimal difference of BC5 and 1DA is equal to 9EB.

这些示例解释了减去两个十六进制数的过程。现在让我们讨论一下十六进制数的第三个基本算术运算，即十六进制乘法。

These examples explain the process of subtracting two hexadecimal numbers. Let us now discuss the third basic arithmetic operation on hexadecimal numbers i.e., hexadecimal multiplications.

Hexadecimal Multiplication

十六进制乘法是一种用于确定两个十六进制数乘积的算术运算。

Hexadecimal multiplication is an arithmetic operation used to determine the product of two hexadecimal numbers.

十六进制乘法类似于十进制乘法。但是，在十六进制乘法的例子中，当乘积大于或等于 16 时，将产生向下一位的进位。

Hexadecimal multiplication is similar to the decimal multiplication. But, in the case of hexadecimal multiplication, a carry is generated to the next column when the product is greater than or equal to 16.

以下示例说明了两个十六进制数相乘的过程。

The following examples demonstrate the process of multiplying two hexadecimal numbers.

Example 1

将 (A19)16 乘以 (B)16。

Multiply (A19)16 by (B)16.

Solution

所给十六进制数的乘积如下所示 −

The multiplication of given hexadecimal numbers is shown below −

(A19)16 乘以 (B)16 = (6F13)16

(A19)16 times (B)16 = (6F13)16

Explanation

将数字 (B)16 与数字 (A19)16 的每位数相乘并写下结果。

Multiply the digit (B)16 with each digit of the number (A19)16 and write down the result.

首先，我们将 B 乘以 9，得到 99 = 96 + 3。因此，3 写为乘积，96 写为进位 6（16 × 6 = 96）至下一位。

Firstly, we multiply B by 9, it gives 99 = 96 + 3. Hence, 3 is written as product and 96 as carry 6 (16 × 6 = 96) to the next column.

然后，我们将 B 乘以 1 并将进位 6 加到乘积中。它得到 17 = 16 + 1。这里，结果是 1，进位是 1。

Then, we multiply B by 1 and add the carry 6 to the product. It gives 17 = 16 + 1. Here, the result is 1 and carry is 1.

最后，我们将 B 乘以 A 并将从上一步得到的进位 1 加到乘积中。它得到 96 + 15（十六进制中的 F）。结果是 F，进位是 6。

Finally, we multiply B by A and add the carry 1 overed from previous step to product. It gives 96 + 15 (F in hexadecimal). The result is F with a carry 6.

因此，A19 和 B 的最终十六进制乘积为 6F13。

Thus, the final hexadecimal product of A19 and B is 6F13.

Example 2

将 (ABC)16 乘以 (29)16。

Multiply (ABC)16 by (29)16.

Solution

所给十六进制数的乘积如下所示 −

The multiplication of given hexadecimal numbers is shown below −

(ABC)16 乘以 (29)16 = (1B81C)16

(ABC)16 times (29)16 = (1B81C)16

Explanation

在此示例中，我们首先将第二个数字 (29)16 的数字 (9)16 乘以数字 (ABC)16 的每一位数字。写下部分乘积。

In this example, we first multiply the digit (9)16 of the second number (29)16 by each digit of the number (ABC)16. Write down the partial product.

然后，我们将数字 (29)16 的数字 (2)16 乘以数字 (ABC)16 的每一位数字。将部分乘积向左移动一位写下来。

Then, we multiply the digit (2)16 of the number (29)16 by each digit of the number (ABC)16. Write down the partial product by shifting one position to the left.

最后，我们将所有部分乘积相加得到最终结果。

Finally, we sum up all the partial products to obtain the final result.

因此，(ABC)16 和 (29)16 的十六进制乘积为 (1B81C)16。

Thus, the hexadecimal product of (ABC)16 and (29)16 is (1B81C)16.

Hexadecimal Division

十六进制除法是我们在基数为 16 的数上执行的第四个基本算术运算。在十六进制除法中，我们得到两个结果，即商和余数。

Hexadecimal division is the fourth basic arithmetic operation that we perform on base-16 numbers. In the hexadecimal division, we obtain two results namely, quotient and remainder.

执行十六进制除法需遵循以下步骤−

The following steps are to be followed to perform the hexadecimal division −

Step 1 − Start diving from the leftmost digit of the dividend.
Step 2 − Multiply the obtained quotient by the divisor and subtract from the dividend.
Step 3 − Bring down the next significant digit or digits of the dividend.
Step 4 − Repeat the process explained in the above three steps until all the digits in the dividend are used.

以下示例演示执行十六进制除法的过程。

The following examples demonstrate the process of performing hexadecimal division.

Example 1

将 (A29)16 除以 (5)16。

Divide (A29)16 by (5)16.

Solution

A29 除以 5 的十六进制除法如下所示 −

The hexadecimal division of A29 by 5 is given below −

在此十六进制除法中，我们得到了商 (208)16 和余数 (1)16。

In this hexadecimal division, we have obtained the quotient (208)16 and remainder (1)16.

Example 2

将 (1AC)16 除以 (A)16。

Divide (1AC)16 by (A)16.

Solution

给定数字的十六进制除法如下所示 −

The hexadecimal division of given numbers is shown below −

在此示例中，我们得到了商 (2A)16 和余数 (8)16。

In this example, we obtained the quotient (2A)16 and the remainder (8)16.

我们解释了执行十六进制算术运算的直接方法。现在，让我们看看如何通过二进制算术执行所有这四个十六进制算术运算。

We explained the direct method of performing hexadecimal arithmetic operations. Now let’s see how we can perform all these four hexadecimal arithmetic operations through binary arithmetic.

正如我们所知，每个十六进制数字可以表示为一组四位二进制数，如下表所示。

As we know, each hexadecimal digit can be represented as group of four bits as shown in the following table.

Hexadecimal	Binary
0	0000
1	0001
2	0010
3	0011
4	0100
5	0101
6	0110
7	0111
8	1000
9	1001
A	1010
B	1011
C	1100
D	1101
E	1110
F	1111

现在，让我们通过二进制转换讨论十六进制算术运算。

Let us now discuss hexadecimal arithmetic operations through binary conversion.

Hexadecimal Addition Through Binary Conversion

在此方法中，通过二进制转换添加两个十六进制数字时，我们首先将它们转换为相应的二进制格式，然后使用二进制算术规则添加二进制数字，最后将最终结果转换回十六进制格式。

In this method of adding two hexadecimal numbers, we first convert them to their equivalent binary format, then add the obtained binary numbers using rules of binary arithmetic, and finally convert the final result back to the hexadecimal format.

以下示例演示通过二进制转换进行十六进制加法的过程。

The following example demonstrates the process of hexadecimal addition through binary conversion.

Example

通过二进制转换添加 (A5C)16 和 (CCD)16。

Add (A5C)16 and (CCD)16 through binary conversion.

Solution

将给定的十六进制数字转换为其二进制等价物，

Converting the given hexadecimal numbers to their binary equivalent,

(A5C)16 = (1010 0101 1100)2

(CCD)16 = (1100 1100 1101)2

将获得的二进制数相加，

Adding the obtained binary numbers,

最后，将二进制和转换成其等效的十六进制格式以获得最终结果。

Finally, converting the binary sum to its equivalent hexadecimal format to obtain the final result.

(0001 0111 0010 1001)2 = (1729)16

因此，(A5C)16 和 (CCD)16 的十六进制和是 (1729)16。

Hence, the hexadecimal sum of (A5C)16 and (CCD)16 is (1729)16.

Hexadecimal Subtraction Through Binary Conversion

要通过二进制转换来减去两个十六进制数，我们首先将给定的十六进制数转换成它们的二进制等效。然后，按照二进制算术规则减去它们。最后，将最终结果转换回十六进制格式。

To subtract two hexadecimal numbers through binary conversion, we first convert the given hexadecimal numbers to their binary equivalent. Then, subtract them as per the rules of binary arithmetic. Finally, convert the final result back to the hexadecimal format.

以下是一个通过二进制转换展示十六进制减法过程的示例。

Here is an example demonstrating the process of hexadecimal subtraction through binary conversion.

Example

使用二进制算术从 (FEA)16 中减去 (AC2)16。

Subtract (AC2)16 from (FEA)16 using binary arithmetic.

Solution

将给定的十六进制数字转换为其二进制等价物，

Converting the given hexadecimal numbers to their binary equivalent,

(FEA)16 = (1111 1110 1010)2

(AC2)16 = (1010 1100 0010)2

减去获得的二进制数，

Subtracting the obtained binary numbers,

将差值转换回十六进制以获得最终结果，

Converting the difference back to the hexadecimal to obtain the final result,

(0101 0010 1000)2 = (528)16

因此，(FEA)16 和 (AC2)16 的十六进制差是 (528)16。

Hence, the hexadecimal difference of (FEA)16 and (AC2)16 is (528)16.

Hexadecimal Multiplication Through Binary Conversion

我们还可以通过将两个十六进制数转换为其二进制等价来相乘。为此，我们首先将给定的十六进制数转换为它们的二进制等价，然后按照二进制算术规则相乘二进制数，并将结果转换回十六进制以获得最终结果。

We can also multiply two hexadecimal numbers by converting them into their binary equivalent. For this, we first convert the given hexadecimal numbers to their binary equivalent, then multiply the binary numbers as per the rules of binary arithmetic, and convert the result back to hexadecimal to obtain the final result.

以下示例展示了通过二进制转换乘以十六进制数的过程。

The following example demonstrates the process of multiplying hexadecimal numbers through binary conversion.

Example

通过二进制转换将 (A9C)16 乘以 (B)16。

Multiply (A9C)16 by (B)16 through binary conversion.

Solution

将给定的十六进制数转换为其等效的二进制，

Converting the given hexadecimal numbers to their equivalent binary,

(A9C)16 = (1010 1001 1100)2

(B)16 = (1011)2

二进制数相乘，

Multiplying the binary numbers,

hexadecimal multiplication binary conversion

将结果转换为十六进制格式，

Converting the product back to the hexadecimal format,

(0111 0100 1011 0100)2 = (74B4)16

因此，A9C和B的十六进制乘积为74B4。

Thus, the hexadecimal product of A9C and B is 74B4.

Hexadecimal Division Through Binary Conversion

十六进制除法也可以使用二进制运算进行。此方法中，首先将给定的十六进制数转换为其二进制等价物，然后再按照二进制除法的规则对其进行除法。最后，我们将结果从二进制格式转换为十六进制以获得最终结果。

The hexadecimal division can also be performed using binary arithmetic. In this method, firstly, we convert the given hexadecimal numbers to their binary equivalent and then divide them as per the rules of binary division. At the end, we convert the result from binary format to hexadecimal to obtain the final result.

让我们借助一个示例了解使用二进制运算的十六进制除法。

Let us understand the hexadecimal division using binary arithmetic with the help of an example.

Example

使用二进制运算将(AB8)16除以(A)16。

Divide (AB8)16 by (A)16 using binary arithmetic.

Solution

将给定的十六进制数字转换为其二进制等价物，

Converting the given hexadecimal numbers to their binary equivalent,

(AB8)16 = (1010 1011 1000)2

(A)16 = (1010)2

对所得二进制数进行除法，

Dividing the obtained binary numbers,

将商和余数转换为十六进制，

Converting the quotient and remainder to hexadecimal,

商 = (0001 0001 0010)2 = (112)16

Quotient = (0001 0001 0010)2 = (112)16

余数 = (0100)2 = (4)16

Remainder = (0100)2 = (4)16

Conclusion

在本章中，我们解释了十六进制数的四种基本算术运算(加法、减法、乘法和除法)。我们还讨论了十六进制算术运算的二进制方法。

In this chapter, we explained the four basic arithmetic operations (addition, subtraction, multiplication, and division) of hexadecimal numbers. We also discussed the binary method of hexadecimal arithmetic operations.