Source code, anti-code, complement, and transfer code

Document directory

• I. Definition of source code, reverse code, and complement code
• Ii. Rules for code addition and subtraction
• I. Floating Point calculation rules

I. Definition of source code, reverse code, and complement code

1. Definition of the original code

① Definition of decimal original code

	[X]Original=		X	0 ≤ x <1
			1-x	-1 <x ≤ 0

Example: X = + 0.1011, [X] original = 01011
X =-0.1011 [X] original = 11011

② Definition of integer original code

	[X]Original=		X	0 ≤ X <2n
			2n-X	-2n <X ≤ 0

2. Definition of supplemental code

① Definition of decimal Complement

	[X]Supplement=		X	0 ≤ X <1
			2 + X	-1 ≤ X <0

Example: X = + 0.1011, [X] fill = 01011
X =-0.1011, [X] fill = 10101

② Integer complement Definition

	[X]Supplement=		X	0 ≤ X <2n
			2n + 1 + X	-2n ≤ X <0

3. Anti-Code definition

① Decimal anticode Definition

	[X]Reverse=		X	0 ≤ X <1
			2-2 n-_x	-1 <X ≤ 0

 

Example: X = + 0.1011 [X] inverse = 01011
X =-0.1011 [X] reversed = 10100

② Integer anticode Definition

	[X]Reverse=		X	0 ≤ X <2n
			2n + 1-_x	-2n <X ≤ 0

4. Shift code: The Shift code is only used to represent the order code of the floating point number, so it is only used as an integer.

① Shift code definition: Set the order code consisting of 1-bit sign bits and n-bit numeric bits, then [X] shift = 2n + X-2n ≤ X ≤ 2n
Example: X = + 1011 [X] shift = 11011 sign bit "1" indicates the positive number
X =-1011 [X] shift = 00101 sign bit "0" indicates negative number

② Relationship between code transfer and CAPTCHA complement: [X] The Relationship Between the CAPTCHA shifting and [X] Is that the signed bits are opposite to each other,
Example: X = + 1011 [X] shift = 11011 [X] fill = 01011
X =-1011 [X] shift = 00101 [X] fill = 10101

③ Considerations for code transfer operations:
◎ The result of the bitwise operation needs to be corrected. The correction amount is 2n, that is, the result is the correct result of the bitwise inversion of the result.
◎ Indicates that 0 has a unique encoding: 1000... 00. When 000... 00 (-2n), which is a floating point.

Ii. Supplementary addition and subtraction operation rules 1. Operation Rules

[X + Y] fill = [X] fill + [Y] fill
[X-Y] fill = [X] fill + [-Y] fill

If the [Y] population is known, the [-Y] population method is to take the [Y] population (including the symbol bit) Bit by bit and then add 1 in the trim bit.
For example, [Y] fill = 101101 [-Y] fill = 010011

2. Overflow judgment. Generally, dual-symbol bit is used for judgment:

Symbol bit 00 indicates positive number 11 indicates negative number
When the result is 01, it is called overflow; when it is 10, it is called underflow.

Example: set x = 0.1101, y =-0.0111, And the symbol bit is double sign bit
Returns x + y and x-y by using the complement code.
[X] fill + [y] fill = 00 1101 + 11 1001 = 00 0110
[X-y] fill = [x] fill + [-y] fill = 00 1101 + 00 0111 = 01 0100
Incorrect result. Positive overflow 3. Implementation of the first-digit multiplication of the original code:

Set X = 0.1101, Y =-0. 1011, and calculate X * Y.
Solution: The symbol bit is processed separately, x + y
The value part uses the original code for a multiplication, as shown in:

	High part product	Low part product/Multiplier			Description
	0 0 0 0 0 0	1 0 1 1			Start Condition
	+) 0 0 1 1 0 1				The multiplier is 1, + X
	0 0 1 1 0 1
	0 0 0 1 1 0	1 1 0 1		1 (lost)	Right shift partial area and multiplier
	+) 0 0 1 1 0 1				The multiplier is 1, + X
	0 1 0 0 1 1
	0 0 1 0 0 1	1 1 1 0		1 (lost)	Right shift partial area and multiplier
	+) 0 0 0 0 0 0				The multiplier percentile is 0, + 0
	0 0 1 0 0 1
	0 0 0 1 0 0	1 1 1 1		0 (lost)	Right shift partial area and multiplier
	+) 0 0 1 1 0 1				The multiplier is 1, + x
	0 1 0 0 0 1
	0 0 1 0 0 0	1 1 1 1		1 (lost)	Right shift partial area and multiplier

4. Implementation of one-digit division of the original code: do not restore the remainder (addition and subtraction alternate method)

	Partial product	Additional bit of the low part product			Operation instructions
	0 0 0 0 0 0	1 0 1 1			Start Condition
	+) 0 0 0 0 0 0				The multiplier is 1, + x
	0 0 0 0 0 0
	0 0 0 0 0 0	1 1 0 1		1 (lost)	Right shift partial area and multiplier
	+) 1 1 0 0 1 1				The multiplier is 1, + X
	0 1 0 0 1 1
	0 0 1 0 0 1	1 1 1 0		1 (lost)	Right shift partial area and multiplier
	+) 0 0 0 0 0 0				The multiplier percentile is 0, + 0
	0 0 1 0 0 1
	0 0 0 1 0 0	1 1 1 1		0 (lost)	Right shift partial area and multiplier
	+) 0 0 1 1 0 1				The multiplier is 1, + X
	0 1 0 0 0 1
	0 0 1 0 0 0	1 1 1 1		1 (lost)	Right shift partial area and multiplier

§ 2. 5 floating point operations and floating point Calculator

I. Floating Point calculation rules 1. Operation steps of floating point addition and subtraction

Set two floating point numbers X = Mx ※2ex Y = My ※2ey
To implement X ± Y, follow these steps:
① Level-1 operation: the level-1 operation is aligned with the level-1 operation.
② Perform addition and subtraction of the number of tails
③ Normalized processing: The result of the operation on the tail number must be a normalized floating point number. For the complement tail number of the double sign bit, it must be
001 ×××... ×× Or 110 ×××... ×× Format
If the Left or Right rules are not met.

④ Rounding operation: When performing the order or right rule operation, you can use the "0" and "1" to round the number of tails that have been removed right to ensure accuracy.
⑤ Correctness of the judgment result: Check whether the level code overflows
If the level code overflows (the shift code indicates that it is 00... 0), set the result to machine 0;
If the level code overflow (exceeds the maximum value indicated by the level code), the overflow flag is set.

Example: Suppose X = 0. 0110011*211, Y = 0.1101101*2-10 (the numbers here are both binary )?? Calculate X + Y;
Solution: [X] float: 0 1 010 1100110
[Y] floating: 0 0 110 1101101
Tail Number of the signed level code

Step 1: Calculate the order difference: │ △e │ = | 1010-0110 | = 0100
Step 2: lower order: the order code of Y is small, and the tail number of Y shifts four places to the right.
[Y] float changes to 0 1 010 0000110 1101 temporarily saved
Step 3: add the ending number and use the double sign bit complement operation
00 1100110
+ 00 0000110
00 1101100
Step 4 Normalization: Meet normalization requirements
Step 5: Use the 0 homes and 1 homes Method for rounding
Therefore, the floating point number format of the final calculation result is 0 1 010 1101101,
That is, X + Y = + 0. 1101101*210

2. Operation steps of floating-point multiplication and division

① Order code operation: sum (multiplication) or difference (Division) of Order code)
That is, [Ex + Ey] shift = [Ex] SHIFT + [Ey] fill
[Ex-Ey] shift = [Ex] SHIFT + [-Ey] fill

② Floating point ending number processing: The result of the floating point number multiplication and division operation must be rounded.
Example: x = 0. 0110011*211, y = 0.1101101*2-10
X ※y
Solution: [x] float: 0 1 010 1100110
[Y] floating: 0 0 110 1101101
Step 1: Add Order Codes
[Ex + ey] shift = [Ex] SHIFT + [ey] fill = 1 010 + 1 110 = 1 000
1 000 is the value 0 indicated by the transfer code
Step 2: the result of multiplying the ending number of the original code is:
0 10101101101110
Step 3: normalization: the specifications have been met, and no left gauge is required. The number of tails remains unchanged and the order code remains unchanged.
Step 4: Rounding: Follow the Rounding Rule and Add 1 to correct
So X ※y = 0.1010111 ※2 + 000