Digital-electronics 简明教程
Digital Electronics - Complement Arithmetic
补码运算是一种在数字电子领域使用的数学技术系统，主要用于执行各种算术运算，即减法。
Complement arithmetic is a system of mathematical techniques used in the field of digital electronics to perform various arithmetic operations mainly, subtraction.
在这里，我们将介绍数字系统中以下最常用的几类补码 −
Here, we will cover the following most widely used types of complements in digital systems −
-
9’s Complement
-
10’s Complement
-
1’s Complement
-
2’s Complement
-
7’s Complement
-
8’s Complement
-
15’s Complement
-
16’s Complement
让我们详细讨论每种补码及其在算术运算中的应用。
Let us discuss each of these complements in detail along with their application in arithmetic operations.
What is 9’s Complement?
在数字电子中，9的补码是一种用于使用数字系统对十进制数执行减法的补码。因此，9的补码与十进制数系统相关。
In digital electronics, the 9’s complement is a type of complement used to perform subtraction of decimal numbers using a digital system. Thus, 9’s complement is related to the decimal number system.
-
9’s complement is used to perform subtraction because it simplifies the subtraction operation.
-
The 9’s complement of a given decimal number is found by subtracting each digit of the number from 9.
下表显示了每个十进制数字的 9 补数 -
The following table shows the 9’s complement of each decimal digit −
Decimal Digit |
9’s Complement |
0 |
9 – 0 = 9 |
1 |
9 – 1 = 8 |
2 |
9 – 2 = 7 |
3 |
9 – 3 = 6 |
4 |
9 – 4 = 5 |
5 |
9 – 5 = 4 |
6 |
9 – 6 = 3 |
7 |
9 – 7 = 2 |
8 |
9 – 8 = 1 |
9 |
9 – 9 = 0 |
让我们借助示例来理解它。
Let us understand it with the help of examples.
Solution
以下是找到给定十进制数的 9 补数的分步过程 -
Here is the step-by-step process for finding 9’s complement of the given decimal number −
-
The 9’s complement of 7 = 9 – 7 = 2
-
The 9’s complement of 8 = 9 – 8 = 1
-
The 9’s complement of 2 = 9 – 2 = 7
-
The 9’s complement of 4 = 9 – 4 = 5
-
The 9’s complement of 4 = 9 – 4 = 5
-
The 9’s complement of 5 = 9 – 5 = 4
-
The 9’s complement of 0 = 9 – 0 = 9
因此，十进制数 7824.450 的 9 补数是 2175.549。
Thus, the 9’s complement of the decimal number 7824.450 is 2175.549.
Solution
十进制数 45608 的 9 补数如下 −
The 9’s complement of the decimal number 45608 is given below −
-
The 9’s complement of 4 = 9 – 4 = 5.
-
The 9’s complement of 5 = 9 – 5 = 4.
-
The 9’s complement of 6 = 9 – 6 = 3.
-
The 9’s complement of 0 = 9 – 0 = 9.
-
The 9’s complement of 8 = 9 – 8 = 1.
因此，45608 的 9 补数是 54391。
Thus, the 9’s complement of 45608 is 54391.
What is 10’s Complement?
在数字电子学中，10 补数是用于执行十进制数减法的另一种类型的补数。同样，10 补数的目的是简化十进制减法运算。
In digital electronics, the 10’s complement is another type of complement used to perform subtraction of decimal numbers. Again, the purpose of the 10’s complement is to simplify the decimal subtraction operation.
有两种求十进制数的 10 补数的方法 −
There are two methods for finding the 10’s complement of a decimal number −
Method I − 要求给定十进制数的 10 补数，首先通过用 9 减去该数的每一位数来求得 9 补数。然后，我们向 9 补数加 1 来获得 10 补数，即，
Method I − To find the 10’s complement of a given decimal number, firstly we find the 9’s complement by subtracting each digit of the number from 9. Then, we add 1 to the 9’s complement to obtain the 10’s complement, i.e.,
10 的补数 = 9 的补数 + 1
10’s Complement = 9’s Complement + 1
使用此方法，每个十进制数字的 10 补数如下表所示 −
The 10’s complement of each decimal digit using this method is given in the following table −
Decimal Digit |
9’s Complement |
0 |
9 – 0 = 9 + 1 = 10 = 0 (Ignore the carry) |
1 |
9 – 1 = 8 + 1 = 9 |
2 |
9 – 2 = 7 + 1 = 8 |
3 |
9 – 3 = 6 + 1 = 7 |
4 |
9 – 4 = 5 + 1 = 6 |
5 |
9 – 5 = 4 + 1 = 5 |
6 |
9 – 6 = 3 + 1 = 4 |
7 |
9 – 7 = 2 + 1 = 3 |
8 |
9 – 8 = 1 + 1 = 2 |
9 |
9 – 9 = 0 + 1 = 1 |
Method II − 本方法中，我们可以使用下列公式找到给定十进制数的十进制补数，
Method II − In this method, we can use the following formula to find the 10’s complement of a given decimal number,
十进制补数 = 10N – 数
10’s Complement = 10N – Number
其中，N 是十进制数中的位数。
Where, N is the number of digits in the decimal number.
让我们通过示例了解求十进制补数的过程。
Let us understand the process of finding the 10’s complement through examples.
Solution
4872 的十进制补数可确定如下 −
The 10’s complement of 4872 can be determined as follows −
求 4872 的九进制补数，
Finding the 9’s complement of 4872,
9999 – 4872 = 5127
对九进制补数加 1 得到十进制补数，
Adding 1 to the 9’s complement to obtain the 10’s complement,
5127 + 1 = 5128
因此，4872 的 10 进制补码是 5128。
So, the 10’s complement of 4872 is 5128.
Solution
2478.98 的各位补数可按照下面得出 −
The 10’s complement of 2478.98 can be found as given below −
找出 2478.98 的九位补数，
Finding the 9’s complement of 2478.98,
9999.99 – 2478.98 = 7521.01
对九进制补数加 1 得到十进制补数，
Adding 1 to the 9’s complement to obtain the 10’s complement,
7521.01 + 1 = 7521.02
因此，7521.01 的各位补数是 7521.02。
Hence, the 10’s complement of 7521.01 is 7521.02.
What is 1’s Complement?
在数字电子学中，一位补数是一种用于简化二进制数运算的补数类型。此外，一位补数用于表示给定二进制数的负数。
In digital electronics, the 1’s complement is a type of complement used to simplify the subtraction of binary numbers. Also, the 1’s complement is used to represent the negative of a given binary number.
我们可以通过将数字中的所有 0 更改为 1，并将所有 1 更改为 0 来找出二进制数的一位补数。
We can find the 1’s complement of a binary number by changing all the 0s to 1s and all the 1s to 0s in the number.
我们还可以通过将数字的每一位从 1 减去来找出二进制数的一位补数。
We can also find the 1’s complement of a binary number by subtracting each bit of the number from 1.
然而，一位补数有一个主要问题，即它对 0 有两种表示形式。其中，00000000 表示正零，其一位补数是 11111111，表示 0，但称为负零。
However, there is a major issue associated with the 1’s complement that is it has two representations for 0. Where, 00000000 represents the positive zero and its 1’s complement is 11111111 that represents 0, but it is called negative zero.
我们来考虑一些示例来理解找出二进制数一位补数的过程。
Let us consider some examples to understand the process of finding the 1’s complement of binary numbers.
Solution
101101 的一位补数可以按如下获得 −
The 1’s complement of 101101 can be obtained as follows −
Method I − 通过翻转每一位 −
Method I − By flipping each bit −
-
The 1’s complement of 1 = 0
-
The 1’s complement of 0 = 1
-
The 1’s complement of 1 = 0
-
The 1’s complement of 1 = 0
-
The 1’s complement of 0 = 1
-
The 1’s complement of 1 = 0
Method II − 将每一位从 1 中减去 −
Method II − By subtract each bit from 1 −
111111 – 101101 = 010010
因此，101101 的 1 的补码为 010010。
Hence, the 1’s complement of 101101 is 010010.
What is 2’s Complement?
在数字电子学中，2 的补码是一个使用数字系统进行二进制减法的概念。
In digital electronics, the 2’s complement is a concept widely used to perform binary subtraction using a digital system.
这里有三种方法可以用来确定给定二进制数的 2 的补码 −
Here are the following three methods that can be used to determine the 2’s complement of a given binary number −
Method I − 通过求出 1 的补码，然后向 1 的补码中添加 1，即
Method I − By finding the 1’s complement and then adding 1 to the 1’s complement, i.e.,
2 的补码 = 1 的补码 + 1
2’s Complement = 1’s Complement + 1
Method II − 通过将给定的二进制数从 2N 中减去，即
Method II − By subtracting the given binary number from 2N, i.e.,
2 的补码 = 2N – 数字
2’s Complement = 2N – Number
其中，“N”是数字中的位数。
Where, "N" is the number of bits in the number.
Method III − 从最低有效位 (LSB) 开始，逐位抄写至遇到第一个 1 位为止，然后对剩下的位求补。
Method III − Starting from the least significant bit (LSB), copy down the bits up to and including the first 1 bit encountered and then complement the remaining bits.
让我们通过实例了解求二进制数的 2 的补码的过程。
Let us understand the process of finding the 2’s complement of binary numbers through examples.
Solution
我们可以按照如下方式找到 1100111 的 2 的补码：
We can find the 2’s complement of 1100111 as follows −
Method I 使用 1 的补码：
Method I − Using 1’s complement −
1100111 的 1 的补码 = 0011000
1’s complement of 1100111 = 0011000
对 1 的补码加 1 以获得 2 的补码，
Adding 1 to 1’s complement to obtain the 2’s complement,
0011000 + 1 = 0011001
Method II 使用 2 的补码公式：
Method II − Using 2’s complement formula −
2 的补码 = 27 – 1100111 = 128 – 1100111
2’s Complement = 27 – 1100111 = 128 – 1100111
2 的补码 = 10000000 – 1100111 = 0011001
2’s Complement = 10000000 – 1100111 = 0011001
Method III 通过从低有效位开始到第一个 1 位复制位：
Method III − By copying down the bits starting from the LSB up to and including the first 1 bit −
Solution
可以按如下方式得到 11001100 的 2 的补码：
The 2’s complement of 11001100 can be obtained as follows −
Method I 使用 1 的补码：
Method I − Using 1’s complement −
11001100 的 1 的补码 = 00110011
1’s complement of 11001100 = 00110011
2 补码 = 1 补码 + 1
2’s complement = 1’s complement + 1
2 的补码 = 00110011 + 1
2’s complement = 00110011 + 1
因此，
Therefore,
2 的补码 = 00110100
2’s complement = 00110100
Method II 通过在 2N 中减去该数字：
Method II − By subtracting the number from 2N −
2 的补码 = 28 - 11001100
2’s complement = 28 - 11001100
2 的补码 = 100000000 – 11001100 = 00110100
2’s complement = 100000000 – 11001100 = 00110100
Method III 通过复制位直到第一个 1 位：
Method III − By copying down bits up to first 1 bit −
What is 7’s Complement?
在数字电子中，7 的补码是一个用于简化八进制减法的概念。给定八进制数的 7 的补码可以通过从 7 中减去该数字的每一位得到。
In digital electronics, the 7’s complement is a concept used to simplify the octal subtraction. The 7’s complement of a given octal number can be obtained by subtracting each digit of the number from 7.
八进制数字的 7 的补数在下面的表格中给出 −
The 7’s complement of each octal digit is given in the following table −
Octal Digit |
7’s Complement |
0 |
7 – 0 = 7 |
1 |
7 – 1 = 6 |
2 |
7 – 2 = 5 |
3 |
7 – 3 = 4 |
4 |
7 – 4 = 3 |
5 |
7 – 5 = 2 |
6 |
7 – 6 = 1 |
7 |
7 – 7 = 0 |
让我们考虑一些例子来理解求八进制数的 7 的补数的过程。
Let us consider some examples to understand the process of finding 7’s complement of an octal number.
Solution
3152 的 7 的补数可按如下方式求得 −
The 7’s complement of 3152 can be obtained as follows −
-
The 7’s complement of 3 = 7 – 3 = 4.
-
The 7’s complement of 1 = 7 – 1 = 6.
-
The 7’s complement of 5 = 7 – 5 = 2.
-
The 7’s complement of 2 = 7 – 2 = 5.
因此，3152 的 7 的补数为 4625。
Hence, the 7’s complement of 3152 is 4625.
What is 8’s Complement?
8 的补数是简化八进制减法而用的另一种补数概念。实际上，它类似于十进制数字系统中的 10 的补数。
The 8’s complement is another type of complement concept used to simplify the octal subtraction. Actually, it is similar to that of 10’s complement in the decimal number system.
我们可以按照以下步骤求出一个给定八进制数的 8 的补数：
We can find the 8’s complement of a given octal number as follows −
-
Find the 7’s complement of the given octal number by subtracting each digit of the number from 7.
-
Add 1 to the 7’s complement.
-
Result will be the 8’s complement of the given octal number.
因此，
Thus,
8 的补数 = 7 的补数 + 1
8’s Complement = 7’s Complement + 1
下表显示了每个八进制数位的 8 的补数：
The following table shows the 8’s complement of each octal digit −
Octal Digit |
8’s Complement |
0 |
7 – 0 = 7 + 1 = 10 = 0 (Ignore the carry) |
1 |
7 – 1 = 6 + 1 = 7 |
2 |
7 – 2 = 5 + 1 = 6 |
3 |
7 – 3 = 4 + 1 = 5 |
4 |
7 – 4 = 3 + 1 = 4 |
5 |
7 – 5 = 2 + 1 = 3 |
6 |
7 – 6 = 1 + 1 = 2 |
7 |
7 – 7 = 0 + 1 = 1 |
让我们借助示例来了解求 8 的补数的过程。
Let us understand the process of finding the 8’s complement with the help of examples.
Solution
4257 的 8 进制补码可以如下找到：
The 8’s complement of 4257 can be found as given below −
4257 的 7 进制补码 = 7777 – 4257 = 3520
7’s complement of 4257 = 7777 – 4257 = 3520
8 进制补码 = 7 进制补码 + 1
8’s complement = 7’s complement + 1
8 进制补码 = 3520 + 1 = 3521
8’s complement = 3520 + 1 = 3521
因此，4257 的 8 进制补码是 3521。
Thus, the 8’s complement of 4257 is 3521.
Solution
给定八进制数的 8 进制补码可以按如下确定：
The 8’s complement of given octal number can be determined as follows −
77201 的 7 进制补码 = 77777 – 77201 = 00576
7’s complement of 77201 = 77777 – 77201 = 00576
8 进制补码 = 7 进制补码 + 1
8’s complement = 7’s complement + 1
8 进制补码 = 00576 + 1 = 00577
8’s complement = 00576 + 1 = 00577
所以，77201 的 8 进制补码是 00577。
So, the 8’s complement of 77201 is 00577.
What is 15’s Complement?
在十六进制数字系统中，15 的补码是用于简化十六进制数字减法运算的一种补码概念。15 的补码类似于十进制数系统中的 9 的补码。
In hexadecimal number system, the 15’s complement is a complement concept used to simplify the subtraction operation of hexadecimal numbers. The 15’s complement is similar to the 9’s complement in decimal number system.
若要找到给定十六进制数的 15 的补码，我们用 15（F）减去该数的每个数字。
To find the 15’s complement of a given hexadecimal number, we subtract each digit of the number from 15 (F).
每个十六进制数字的 15 的补码在以下表格中给出：
The 15’s complement of each hexadecimal digit is given in the following table −
Hexadecimal Digit |
15’s Complement |
0 |
F – 0 = F |
1 |
F – 1 = E |
2 |
F – 2 = D |
3 |
F – 3 = C |
4 |
F – 4 = B |
5 |
F – 5 = A |
6 |
F – 6 = 9 |
7 |
F – 7 = 8 |
8 |
F – 8 = 7 |
9 |
F – 9 = 6 |
A |
F – A = 5 |
B |
F – B = 4 |
C |
F – C = 3 |
D |
F – D = 2 |
E |
F – E = 1 |
F |
F – F = 0 |
以下示例展示了查找十六进制数的 15 的补数的过程。
The following examples demonstrate the process of finding 15’s complement of the hexadecimal numbers.
Solution
A259C 的 15 的补数可以按如下方式获取 −
The 15’s complement of A259C can be obtained as follows −
-
The 15’s complement of A = F – A = 5.
-
The 15’s complement of 2 = F – 2 = D.
-
The 15’s complement of 5 = F – 5 = A.
-
The 15’s complement of 9 = F – 9 = 6.
-
The 15’s complement of C = F – C = 3.
因此，A259C 的 15 进制补码是 5DA63。
So, the 15’s complement of A259C is 5DA63.
What is 16’s Complement?
在十六进制算术中，我们还可以确定给定十六进制数的 16 进制补码。16 进制补码是一个用于简化十六进制数减法运算的概念。
In hexadecimal arithmetic, we can also determine the 16’s complement of a given hexadecimal number. The 16’s complement is a concept used to simplify the subtraction operation of hexadecimal numbers.
我们可以按如下所述确定给定十六进制数的 16 进制补码：
We can determine the 16’s complement of a given hexadecimal number as described below −
-
Find the 15’s complement of the given hexadecimal number.
-
Add 1 to the 15’s complement obtained. This gives the 16’s complement of the hexadecimal number.
因此，
Therefore,
16 进制补码 = 15 进制补码 + 1
16’s Complement = 15’s Complement + 1
每个十六进制数字的 16 进制补码在以下表中给出：
The 16’s complement of each hexadecimal digit is given in the following table −
Hexadecimal Digit |
15’s Complement |
0 |
F – 0 = F + 1 = 10 = 0 (Ignore the carry) |
1 |
F – 1 = E + 1 = F |
2 |
F – 2 = D + 1 = E |
3 |
F – 3 = C + 1 = D |
4 |
F – 4 = B + 1 = C |
5 |
F – 5 = A + 1 = B |
6 |
F – 6 = 9 + 1 = A |
7 |
F – 7 = 8 + 1 = 9 |
8 |
F – 8 = 7 + 1 = 8 |
9 |
F – 9 = 6 + 1 = 7 |
A |
F – A = 5 + 1 = 6 |
B |
F – B = 4 + 1 = 5 |
C |
F – C = 3 + 1 = 4 |
D |
F – D = 2 + 1 = 3 |
E |
F – E = 1 + 1 = 2 |
F |
F – F = 0 + 1 = 1 |
让我们举些例子来理解求十六进制数的16进制补码的过程。
Let us take some examples to understand the process of finding the 16’s complement of hexadecimal numbers.
Solution
给定的十六进制数的16进制补码可按如下方式确定−
The 16’s complement of the given hexadecimal number can be determined as follows −
1ABDF7的15进制补码= FFFFFF-1ABDF7=E54208
15’s complement of 1ABDF7 = FFFFFF – 1ABDF7 = E54208
16进制补码= 15进制补码+1
16’s complement = 15’s complement + 1
16进制补码= E54208+1=E54209
16’s complement = E54208 + 1 = E54209
因此，1ABDF7的16进制补码是E54209。
Thus, the 16’s complement of 1ABDF7 is E54209.
Solution
ABC的16进制补码为，
The 16’s complement of ABC is,
ABC的15进制补码= FFF-ABC=543
15’s complement of ABC = FFF – ABC = 543
16进制补码= 15进制补码+1
16’s complement = 15’s complement + 1
16进制补码= 543+1=544
16’s complement = 543 + 1 = 544
因此，ABC 的 16’s 补码是 544。
Thus, the 16’s complement of ABC is 544.
这些都是关于在数字电子产品中使用的不同类型的补码。
This is all about finding different types of complements used in digital electronics.
现在，让我们看看它们在执行减法运算中的应用。
Now, let us see their application in performing subtraction operations.
Subtraction using 9’s Complement
9’s 补码可用于执行十进制数的减法。在此方法中，通过将被减数的 9’s 补码加到被减数中得到两个十进制数的差值。
The 9’s complement can be used to perform subtraction of decimal numbers. In this method, the difference of two decimal numbers is obtained by adding the 9’s complement of the subtrahend to the minuend.
让我们通过一个示例来了解使用 9’s 补码进行减法。
Let us understand the subtraction using 9’s complement through an example.
Solution
在此示例中，我们有：
In this example, we have,
被减数 = 729
Minuend = 729
减数 = 517
Subtrahend = 517
求出 517 的 9’s 补码，得到：
Finding the 9’s complement of 517, we get
999 – 517 = 482
现在，将 729 和 482 相加得到 729 和 517 的差值，得到：
Now, adding 729 and 482 to obtain the difference of 729 and 517, we get,
729 + 482 = 1211
有环形进位，指示结果为正，并且通过将环形进位加到中间结果的 LSD 中得到最终结果，即：
There is an end around carry, indicating the result is positive and is obtained by adding the end around carry to the LSD of intermediate result to obtain the final result, i.e.,
211 + 1 = 212
因此，729 和 517 的差值为 212。
So, the difference of 729 and 517 is 212.
Example 2
使用 9’s 补码方法从 (159)10 中减去 (203)10。
Subtract (203)10 from (159)10 using 9’s complement method.
Solution
在此示例中，
In this example,
被减数 = 159
Minuend = 159
减数 = 203
Subtrahend = 203
999 减去 203，得到：
Taking 9’s complement of 203, we get,
999 – 203 = 796
将 159 和 796 相加，得到：
Adding 159 and 796, we get,
159 + 796 = 955
没有环绕进位。因此，最终结果为负，可通过取 955 的 9 的补数得到，即：
There is no end around carry. Thus, the final result is negative and obtained by taking the 9’s complement of 955, i.e.,
999 – 955 = 44
因此，减法 159 – 203 的最终结果为 –44。
Thus, the final result of subtraction 159 – 203 = –44.
Subtraction using 10’s Complement
我们还可以使用 10 的补数进行十进制减法。下面给出了使用 10 的补数执行十进制减法的分步过程 −
We can also perform decimal subtraction using 10’s complement. The step-by-step procedure to perform decimal subtraction using 10’s complement is given below −
-
Step 1 − Consider the decimal subtraction, X – Y. Where, X is minuend and Y is subtrahend.
-
Step 2 − Find 10’s complement of Y.
-
Step 3 − Add X and 10’s complement of Y.
-
Step 4 − If there is an end-around carry, the result will be positive and final result is obtained by discarding the carry. If there is no end-around carry, it indicates that the result is negative and is obtained by taking 10’s complement of the intermediate result and assign a negative sign before it.
让我们通过示例了解使用 10 的补数进行十进制减法。
Let us understand the decimal subtraction using 10’s complement with the help of examples.
Example 1
使用 10 的补数算术从 (875)10 中减去 (599)10。
Subtract (599)10 from (875)10 using 10’s complement arithmetic.
Solution
在此示例中，我们有：
In this example, we have,
被减数 = 875
Minuend = 875
减数 = 599
Subtrahend = 599
求 599 的 10 的补数，得到：
Finding 10’s complement of 599, we get,
599 的 10 的补数 = 9 的补数 + 1
10’s complement of 599 = 9’s complement + 1
因此，
Therefore,
599 的 10 的补数 = (999 – 599) + 1 = 401
10’s complement of 599 = (999 – 599) + 1 = 401
875 和 401 相加，我们得到：
Adding 875 and 401, we get,
875 + 401 = 1276
有端部进位，表示结果为正数并且是通过丢弃进位获得的。
There is an end-around carry, showing that the result is positive and is obtained by discarding the carry.
因此，875 和 599 的差为 276。
Thus, the difference of 875 and 599 is 276.
Example 2
使用 10 进制补码运算从 (279)10 中减去 (307)10。
Subtract (307)10 from (279)10 using 10’s complement arithmetic.
Solution
我们有：
We have,
被减数 = 279
Minuend = 279
减数 = 307
Subtrahend = 307
求 307 的 10 进制补码，我们得到：
Taking 10’s complement of 307, we get,
307 的 10 进制补码 = (999 - 307) + 1 = 693
10’s complement of 307 = (999 – 307) + 1 = 693
将 279 和 693 相加，我们得到：
Adding 279 and 693, we get,
279 + 693 = 972
没有端部进位，表示结果为负数。最终结果是求 972 的 10 进制补码，即：
There is no end-around carry, indicating that the result is negative. The final result is obtained by taking the 10’s complement of 972 i.e.,
972 的 10 进制补码 = (999 - 972) + 1 = 28
10’s complement of 972 = (999 - 972) + 1 = 28
因此，最终结果为 -28。
Hence, the final result is –28.
Subtraction using 1’s Complement
1 进制补码用于二进制减法运算。
The 1’s complement is used in binary subtraction operation.
两个二进制数的减法，例如 X 和 Y，即 X - Y，可以使用 1 进制补码按以下步骤执行：
The subtraction of two binary numbers, say X and Y i.e., X – Y, can be performed using 1’s complement as per the following steps −
-
Step 1 − Find the 1’s complement of subtrahend (Y).
-
Step 2 − Add X and 1’s complement of Y.
-
Step 3 − If there is an end-around carry, it indicates the result is positive and final result is obtained by adding the end-around carry to the LSB of the intermediate result. If there is no end-around carry, the result is negative and obtained by taking 1’s complement of the intermediate result and put a negative sign in front of it.
考虑以下示例来理解使用 1 的补码算术的二进制减法。
Consider the following examples to understand the binary subtraction using 1’s complement arithmetic.
Solution
在此示例中，我们有：
In this example, we have,
被减数 = 1011
Minuend = 1011
减数 = 0111
Subtrahend = 0111
查找 1 的补码减数，
Finding 1’s complement of subtrahend,
0111 的 1 的补码 = 1000
1’s complement of 0111 = 1000
将 1011 和 1000 相加，得到，
Adding 1011 and 1000, we get,
1011 + 1000 = 1 0011
存在环绕进位，表示结果为正。最终结果是通过将此环绕进位加到中间结果 (0011) 的 LSB，即获得的，
There is an end-around carry, indicating that the result is positive. The final result is obtained by adding this end-around carry to LSB of the intermediate result (0011) i.e.,
0011 + 1 = 0100
因此，1011 和 111 的二进制差为 100。
Hence, the binary difference of 1011 and 111 is 100.
Example 2
使用 1 的补码算术从 (111)2 中减去 (1100)2。
Subtract (1100)2 from (111)2 using 1’s complement arithmetic.
Solution
我们得到，
We are given,
被减数 = 0111
Minuend = 0111
减数 = 1100
Subtrahend = 1100
查找 1 的补码减数，
Finding the 1’s complement of the subtrahend,
1100 的 1 的补码 = 0011
1’s complement of 1100 = 0011
将 0111 和 0011 相加，得到，
Adding 0111 and 0011, we get,
0111 + 0011 = 1010
没有环绕进位，表示该结果为负，并且通过取 1010 的 1 的补码获得，即：
There is no end-around carry, indicating that the result is negative and is obtained by taking 1’s complement of 1010 i.e.,
1010 的 1 的补码 = 0101
1’s complement of 1010 = 0101
因此，111 和 1100 的二进制差为 -101。
Hence, the binary difference of 111 and 1100 is -101.
Subtraction using 2’s Complement
2 的补码也用于使用数字系统执行二进制减法运算。执行两个二进制数（例如 X 和 Y）的减法、即 (X – Y) 的步骤如下：
The 2’s complement is also used to perform binary subtraction operation using digital systems. The step-by-step procedure to perform the subtraction of two binary numbers, say X and Y, i.e., (X – Y) is given as follows −
-
Step 1 − Find the 2’s complement of subtrahend (Y).
-
Step 2 − Add X and 2’s complement of Y.
-
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by ignoring the end-around carry. If there is no end-around carry, the result is negative and obtained by taking 2’s complement of the intermediate result and put a negative sign in front of it.
让我们看一些示例来了解使用 2 的补码运算法则执行二进制减法。
Let us see some examples to understand binary subtraction using 2’s complement arithmetic.
Example 1
使用 2 的补码运算法则从 (1100)<sub>2</sub>中减去 (101)<sub>2</sub>。
Subtract (101)2 from (1100)2 using 2’s complement arithmetic.
Solution
此示例中，我们所给的是：
In this example, we are given,
被减数 = 1100
Minuend = 1100
减数 = 0101
Subtrahend = 0101
取减数的 2 的补码，我们得到：
Taking 2’s complement of the subtrahend, we get,
0101 的 2 的补码 = (1111 - 0101) + 1 = 1011
2’s complement of 0101 = (1111 - 0101) + 1 = 1011
添加 1100 和 1011，我们得到：
Adding 1100 and 1011, we get,
1100 + 1011 = 1 0111
存在环绕进位，表示结果为正且最终结果通过忽略此环绕进位获得。
There is an end-around carry, indicating that the result is positive and the final result is obtained by ignoring this end around carry.
因此，1100 和 101 的二进制差为 111。
Thus, the binary difference of 1100 and 101 is 111.
Example 2
使用 2 的补码算法从 (0110)2 中减去 (1010)2。
Subtract (1010)2 from (0110)2 using 2’s complement arithmetic.
Solution
给定的数字为：
The given numbers are,
被减数 = 0110
Minuend = 0110
减数 = 1010
Subtrahend = 1010
取减数的 2 的补码，我们得到：
Taking 2’s complement of the subtrahend, we get,
1010 的 2 的补码 = ((1111 - 1010) + 1) = 0110
2’s complement of 1010 = (1111 - 1010) + 1 = 0110
被减数与减数的 2 的补码相加，得到：
Adding minuend and 2’s complement of subtrahend, we get,
0110 + 0110 = 1100
由于没有环绕进位，表明结果为负数。最终结果通过求中间结果的 2 的补码得到，即：
Since, there is no end-around carry, indicating that the result is negative. The final result is obtained by taking 2’s complement of intermediate result, i.e.,
1100 的 2 的补码 = ((1111 - 1100) + 1) = 0100
2’s complement of 1100 = (1111 - 1100) + 1 = 0100
因此，0110 和 1010 的二进制差为 -100。
Thus, the binary difference of 0110 and 1010 is -100.
Subtraction using 7’s Complement
7 的补码算法可用于执行八进制数的减法。以下是使用 7 的补码执行八进制减法的步骤：
The 7’s complement arithmetic can be used to perform subtraction of octal numbers. Here are the steps involved in performing octal subtraction using 7’s complement.
令我们要从 X 中减去八进制数 Y，即 X – Y，则
Let we want to subtract octal number Y from X i.e., X – Y, then
-
Step 1 − Find the 7’s complement of subtrahend (Y).
-
Step 2 − Add X and 7’s complement of Y.
-
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by adding the end-around carry to the intermediate result. If there is no end-around carry, the result is negative and is obtained by taking 7’s complement of the intermediate result and put a negative sign in front of it.
让我们了解一下使用 7 的补码算法的八进制减法。
Let us understand octal subtraction using 7’s complement arithmetic.
Example 1
使用 7 的补码算法从 (721)8 中减去 (540)8。
Subtract (540)8 from (721)8 using 7’s complement arithmetic.
Solution
给定的数字为：
Given numbers are,
被减数 = 721
Minuend = 721
被减数 = 540
Subtrahend = 540
取被减数的 7 补码，
Taking 7’s complement of the subtrahend,
540 的 7 补码 = 777 – 540 = 237
7’s complement of 540 = 777 – 540 = 237
加 7 补码和被减数，得，
Adding 7’s complement and minuend, we get,
721 + 237 = 1 160
有进位环绕，表明结果为正，最终结果通过将进位环绕加到中间结果，来得到，即
There is an end-around carry, indicating that the result is positive and the final result is obtained by adding this end-around carry to the intermediate result, i.e.,
160 + 1 = 161
因此，721 和 540 的八进制差是 161。
Thus, the octal difference of 721 and 540 is 161.
Solution
在这个例子中，给定的数字是，
In this example, the given numbers are,
被减数 = 121
Minuend = 121
被减数 = 310
Subtrahend = 310
取被减数的 7 补码，得，
Taking 7’s complement of the subtrahend, we get,
310 的 7 补码 = 777 – 310 = 467
7’s complement of 310 = 777 – 310 = 467
加被减数和被减数的 7 补码，即，
Adding minuend and the 7’s complement of the subtrahend, i.e.,
121 + 467 = 610
由于没有进位环绕，表明结果为负，可以通过获取中间结果的 7 补码来得到，即，
Since, there is no end around carry, indicating the result is negative and is obtained by taking 7’s complement of the intermediate result i.e.,
610 的 7 补码 = 777 – 610 = 167
7’s complement of 610 = 777 – 610 = 167
因此，121 和 310 的八进制差是 -167。
Hence, the octal difference of 121 and 310 is -167.
Subtraction using 8’s Complement
8 的补数是另一个用于执行八进制减法的技术。下面解释了使用 8 的补数执行八进制减法的逐步过程 −
The 8’s complement is another technique that used to perform octal subtraction. The step-by-step procedure to perform octal subtraction using 8’s complement is explained below −
-
Step 1 − If an octal subtraction is defined as X – Y. Then, find the 8’s complement of subtrahend (Y).
-
Step 2 − Add X and 8’s complement of Y.
-
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by ignoring the end-around carry. If there is no end-around carry, the result is negative and is obtained by taking the 8’s complement of the intermediate result and put a negative sign in front of it.
让我们通过示例了解使用 8 的补数方法的八进制减法。
Let us understand the octal subtraction using 8’s complement method through examples.
Example 1
使用 8 的补数算法从 (712)8 中减去 (103)8。
Subtract (103)8 from (712)8 using 8’s complement arithmetic.
Solution
我们得到，
We are given,
被减数 = 712
Minuend = 712
减数 = 103
Subtrahend = 103
找出减数的 8 的补数，我们得到，
Finding the 8’s complement of subtrahend, we get,
103 的 8 的补数 = (777 - 103) + 1 = 675
8’s complement of 103 = (777 - 103) + 1 = 675
将被减数和减数的 8 的补数相加，我们得到，
Adding the minuend and 8’s complement of subtrahend, we get,
712 + 675 = 1607
存在进位。最终结果通过忽略进位获得。
There is an end-around carry. The final result is obtained by ignoring the end-around carry.
因此，712 和 103 的八进制差为 607。
Thus, the octal difference of 712 and 103 is 607.
Example 2
使用 8 的补数方法从 (206)8 中减去 (471)8。
Subtract (471)8 from (206)8 using the 8’s complement method.
Solution
在此示例中，我们有：
In this example, we have,
被减数 = 206
Minuend = 206
减数 = 471
Subtrahend = 471
找出减数的 8 的补数，
Finding the 8’s complement of subtrahend,
471 的 8 的补数 = (777 - 471) + 1 = 307
8’s complement of 471 = (777 - 471) + 1 = 307
将被减数和减数的 8 的补数相加，我们得到，
Adding the minuend and 8’s complement of subtrahend, we get,
206 + 307 = 515
由于没有环绕进位，因此最终结果为负数，可通过获取中间结果的 8 进制补码来获得，即：
Since, there is no end-around carry, hence the final result is negative and is obtained by taking the 8’s complement of the intermediate result i.e.,
515 的 8 进制补码 = (777 - 515) + 1 = 263
8’s complement of 515 = (777 - 515) + 1 = 263
因此，206 和 471 的 8 进制差为 -263。
Hence, the octal difference of 206 and 471 is -263.
Subtraction using 15’s Complement
15 的补码用于执行十六进制数的减法。如果我们要从 X 中减去十六进制数 Y，则遵循以下步骤：
The 15’s complement is used to perform subtraction of hexadecimal numbers. If we want to subtract a hexadecimal number Y from X, then we follow the steps given below −
-
Step 1 − Find the 15’s complement of subtrahend (Y).
-
Step 2 − Add X and 15’s complement of Y.
-
Step 3 − If there is an end-around carry, it shows the result is positive and the final result is obtained by adding the end-around carry to the intermediate result. If there is no end-around carry, the result is negative and is obtained by taking 15’s complement of the intermediate result and put a negative sign in front of it.
以下示例演示了使用 15 的补码算术执行十六进制减法的过程。
The following examples demonstrate the process of performing hexadecimal subtraction using 15’s complement arithmetic.
Example 1
使用 15 的补码算术从 (E57A)16 中减去 (1920)16。
Subtract (1920)16 from (E57A)16 using 15’s complement arithmetic.
Solution
给定的数字为：
The given numbers are,
被减数 = E57A
Minuend = E57A
减数 = 1920
Subtrahend = 1920
求出减数的 15 的补码：
Finding 15’s complement of the subtrahend,
减数的 15 的补码 = FFFF – 1920 = E6DF
15’s complement of subtrahend = FFFF – 1920 = E6DF
将被减数和减数的 15 的补码相加，得到：
Adding the minuend and 15’s complement of subtrahend, we get,
E57A + E6DF = 1 CC59
存在环绕进位，表明结果为正数，可通过将环绕进位添加到中间结果中得到，即：
There is an end around carry showing that the result is positive and is obtained by adding the end-around carry to the intermediate result, i.e.,
CC59 + 1 = CC5A
因此，E57A 和 1920 的十六进制差为 CC5
Hence, the hexadecimal difference of E57A and 1920 is CC5A.
Example 2
使用 8 的补码运算，从 (A209)16 中减去 (DC25)16。
Subtract (DC25)16 from (A209)16 using 8’s complement arithmetic.
Solution
给定的数字为：
Given numbers are,
被减数 = A209
Minuend = A209
减数 = DC25
Subtrahend = DC25
求减数的 15 的补码，
Finding 15’s complement of subtrahend,
DC25 的 15 的补码 = FFFF - DC25 = 23DA
15’s complement of DC25 = FFFF – DC25 = 23DA
相加被减数和减数的 15 的补码，得到，
Adding minuend and 15’s complement of subtrahend, we get,
A209 + 23DA = C5E3
由于没有环绕进位。结果为负数，且可通过取中间结果的 15 的补码得到，即
Since, there is no end-around carry. The result is negative and is obtained by taking 15’s complement of intermediate result, i.e.,
C5E3 的 15 的补码 = FFFF - C5E3 = 3A1C
15’s complement of C5E3 = FFFF – C5E3 = 3A1C
因此，A209 和 DC25 的十六进制差为 -3A1C。
Hence, the hexadecimal difference of A209 and DC25 us -3A1C.
Subtraction using 16’s Complement
16 的补码也可用于执行十六进制减法。以下是使用 16 的补码进行十六进制减法的步骤：
The 16’s complement is also used to perform hexadecimal subtraction. The steps involved in hexadecimal subtraction using 16’s complement is explained here −
-
Step 1 − If a hexadecimal subtraction is defined as X – Y. Then, find the 16’s complement of subtrahend (Y).
-
Step 2 − Add X and 16’s complement of Y.
-
Step 3 − If there is an end-around carry, it indicates the result is positive and the final result is obtained by ignoring the end-around carry. If there is no end-around carry, the result is negative and is obtained by taking the 16’s complement of the intermediate result and put a negative sign in front of it.
让我们看一些示例，以了解使用 16 的补码进行十六进制减法。
Let us see some examples to understand the hexadecimal subtraction using 16’s complement.
Example 1
使用 16 的补码运算，从 (F9D)16 中减去 (E7C)16。
Subtract (E7C)16 from (F9D)16 using 16’s complement arithmetic.
Solution
所给的十六进制数为
The given hexadecimal numbers are
被减数 = F9D
Minuend = F9D
减数 = E7C
Subtrahend = E7C
通过对被减数进行 16 进制补码计算，我们得到，
Taking 16’s complement of the subtrahend, we get,
E7C 的 16 进制补码 = (FFF – E7C) + 1 = 184
16’s complement of E7C = (FFF – E7C) + 1 = 184
对被减数和被减数的 16 进制补码进行加法计算，我们得到，
Adding minuend and 16’s complement of subtrahend, we get,
F9D + 184 = 1121
产生了循环进位，表示结果为正数。通过忽略循环进位即可获得最终结果。
There is an end-around carry, indicating the result is positive. The final result is obtained ignoring the end-around carry.
因此，F9D 和 E7C 的十六进制差为 121。
Hence, the hexadecimal difference of F9D and E7C is 121.
Example 2
通过 16 进制补码法从 (AC5)16 中减去 (FF2)16。
Subtract (FF2)16 from (AC5)16 using 16’s complement method.
Solution
给定的数字为：
The given numbers are,
被减数 = AC5
Minuend = AC5
被减数 = FF2
Subtrahend = FF2
通过对被减数进行 16 进制补码计算，我们得到，
Taking the 16’s complement of the subtrahend, we get,
FF2 的 16 进制补码 = (FFF – FF2) + 1 = 00E
16’s complement of FF2 = (FFF – FF2) + 1 = 00E
对被减数和被减数的 16 进制补码进行加法计算，我们得到，
Adding the minuend and 16’s complement of subtrahend, we get,
AC5 + 00E = AD3
没有产生循环进位，表示结果为负数。通过对中间结果进行 16 进制补码计算，得到最终结果，如下所示 −
There is no end-around carry, indicating that the result is negative. The final result is obtained by taking the 16’s complement of the intermediate result, as follows −
AD3 的 16 进制补码 = (FFF – AD3) + 1 = 52D
16’s complement of AD3 = (FFF – AD3) + 1 = 52D
因此，AC5 和 FF2 的十六进制差为 -52D。
Hence, the hexadecimal difference of AC5 and FF2 is -52D.
Conclusion
总结来说，补码运算是一种在数字电子产品中用于简化减法运算的方法。在本章中，我们解释了不同类型的补码及其在减法运算中的应用，并给出了已解决的示例。
In conclusion, complement arithmetic is a method used in digital electronics for simplifying subtraction operations. In this chapter, we explained the different types of complements and their application in subtraction operations along with solved examples.