C language - vernacular to understand the original code, inverse code and complement code

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Original, Inverse and Complement

1. Understand storage types

1. Data is divided into byte type and word type data
   Byte type data occupies 8 bits, its original code, inverse code and complement code are all 8 bits, indicating 0-255
   Font type data occupies 8 bits16 bits, its original code, inverse code and complement code are all 16 bits, which means 0-65535
2. Then some people will have doubts: 8-bit binary and 16-bit binary are both 1. Of course, it can represent 255 or 65535, but I can also represent -127, -32767.Now that we say that, do we have to figure out what storage method is, whether it is unsigned storage or signed storage?
3. In 8 bits, if it is unsigned storage - that is, 0-255 has no negative number, if it is a signed number, it will occupy one bit by itself, so it is represented as -127-128.
   So I know that our C language int type is also a signed number when it is not written unsigned, its range is -2147483648~+2147483647, and the signed number is 4,294,967,295
4. So what is the source code stored?Are we talking about the original code, whether it is signed or unsigned?It is very responsible to tell you that this is a signed code, so we look at the original code of 8 digits or the original code of 16 digits, and we can see whether it is positive or negative at a glance

2. Original code, inverse code, complement code

1. Throwing out the concept of three codes

Original code: the highest bit is the sign bit, 0 represents a positive number, 1 represents a negative number, and the others are 0 or 1.Inverse code: The inverse code of a positive number is consistent with the original code, and the inverse code of a negative number is the inverse of the original code, and the sign bit remains unchanged.Complement code: The complement code of a positive number is the same as the original code, and the complement code of a negative number is to invert the original code and add 1, and the sign bit remains unchanged.

Example00000001 (Decimal from the original code is 1)Example 10000001 (The decimal derived from the original code is -1??)So is it?00000001+ 10000001------------100000101+(-1) = -2??Outrageous!!!!

1. But I tell you responsibly, this is -1, but if you count it directly, it's definitely not like this.To perform binary addition and subtraction, it finally becomes addition.
   Do we have a question, the computer -1+1 is not equal to 0 directly, so what should the computer do, how can it be equal to 0?
2. We want to find out 1+(-1)=0 and make it true, then we assume that -1 has a substitute, which is an eight-digit unknown x

 00000001+ xxxxxxxx------------00000000

1. It can be obtained that if 11111111 is added to 1 and becomes 100000000, it will cause a bit overflow and lose the highest bit 1, so isn't this a decimal 0?So I'm sure that adding (-1) equals 0, that is, adding his substitute, he is 111111111
2. It turned out to be like this, except for the highest bit, I invert all of them and then +1
   This is also the case, the concept of complement code is also the same, that is, the original code except the highest bit is inverted and then +1, however his positive calculation process is

Example 10000001 original codeIf you want to change the complement code, you can change the inverse code before you can , inverse bit by bit, except for the sign bit of the highest bitExample 11111110 inverse codeTo change to complement code, just add 1 after the inverse code.Example 11111111 complement (original code except the highest bit is inverted and then +1)

So, in a sense, this becomes 1 + (-127) = 0, but don't confuse it, it will be stressful to understand
We still understand it honestly: for calculationBoth are calculated as the complement of positive or negative numbers.

2. Final summary

The purpose of introducing the inverse code and the complement code of the original code is to solve the problem of subtraction, because there is only an adder in the arithmetic unit of the computer CPU, and the subtraction must be converted into an addition for calculation.

3. To Readers

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