Registers

Internal registers are containers of data. Data values of registers can be modified by a series of register-related instructions.

0x10 Write Value of a Register

Modify value in a register.

tip

Values of all registers are available on page 4 to page 7 on secondary display.

INST_WRITE_REG	HEX	sOP00	sOP01	OP-1
Write 03E8 to R1C	10	00	1C	03E8

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Target register
OP-1	0000~FFFF	Value

0x11 Register Copy

Copy value of one register to another.

tip

Check the result on the secondary display on the corresponding page.

INST_REG_COPY	HEX	sOP00	sOP01	sOP10	sOP11
Copy value from R1C to R1D	11	00	1D	00	1C

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Target register Index
sOP10	00	Ignored
sOP11	00~3F	Source register Index

0x12 Increment/Decrement

Increment or decrement register value.

tip

Rules for overflow can be applied in this instruction.

caution

If overflow is not allowed, the maximum value of a register is 0xFFFF when keep on increasing, and the minimum value is 0x0000 when keep on decreasing.

INST_REG_INC_DEC	HEX	sOP00	sOP01	OP-1
Increase R1C by 03E8，overflow not permitted	12	00	1C	03E8
Increase R1C by 0FFF，overflow permitted	12	01	1C	0FFF

Suppose the value in R1C is FF00, in the first example, the increment is 03E8:

$$(FF00)_{16}+(03E8)_{16}=(1\;02E8)_{16}$$

The result should be 1 02E8, which needs 17 bits. As overflow is not allowed as configured in the opcode, only the least 16 bits will be kept in the register R1C, leaving 0x02E8 as the result.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Increasing, overflow NOT allowed
	01	Increasing, overflow allowed
	10	Decreasing, overflow NOT allowed
	11	Decreasing, overflow allowed
sOP01	00~3F	Target register index
OP-1	00~FF	Value to increase or decrease

0x13 Register Value to Digital Output

Use the least 4 bits are digital output.

INST_REG_TO_PORT	HEX	sOP00	sOP01	OP-1
Use the least 4 bits of R04 as digital output value	13	00	04	0000

Suppose the current value in R04 is 3579:

$$(3579)_{16}=(13689)_{10}$$

In binary:

$$(11\;0101\;0111\;1001)_{2}$$

The last 4 bits are:

$$(9)_{16}=(9)_{10}=(1001)_{2}$$

The hexadecimal value 9, or b1001 in binary, will be sent to the digital output.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Source register index
OP-1	0000	Ignored

0x14 Digital Input Value to Register

Capture the input values on the 4 digital input pins and save it to a register.

tip

The digital input input value ranges in 0000~000F.

$$(000F)_{16}=(1111)_{2}$$

The maximum value is 0xF when all bits are 1.

INST_PORT_TO_REG	HEX	sOP00	sOP01	OP-1
Save the digital input value in R04	14	00	04	0000

Suppose the four pins are HIGH, LOW, HIGH and LOW, i.e. 1010, then:

$$(1010)_{2}=(000C)_{16}$$

In hexadecimal it is 000C, which is saved in R04.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Target register index
OP-1	0000	Ignored

0x15 Logic Operation: Bit Shift

Shift the bits to left or to right in a register for specified bit positions, in circular mode or not.

INST_BITSHIFT	HEX	sOP00	sOP01	sOP10	sOP11
Shift R07 2 bits to the right	15	00	07	00	02
Shift R07 2 bits to the left, in circular mode	15	11	07	00	02

Suppose the current value in R07 is 8009, in binary, it is:

$$(8009)_{16}=(1000\;0000\;0000\;1001)_{2}$$

If circular mode is not allowed, shifting 2 bits to the right:

$$(0010\;0000\;0000\;0010)_{2}=(2002)_{16}$$

The least 2 bits are shifted out, therefore it become 2002; in circular mode, it is:

$$(0110\;0000\;0000\;0010)_{2}=(6002)_{16}$$

The least 2 bits 01 are shifted out and shifted in from the left, therefore it become 6002.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Shift to the right
	01	Shift to the right in circular mode
	10	Shift to the left
	11	Shift to the left in circular mode
sOP01	00~3F	Register index
sOP10	00	Ignored
sOP11	00~FF	Number of bits to shift

0x16 Logic Operation: Bits Inversion

Inverse the bits in a register.

INST_BIT_REVERSE	HEX	sOP00	sOP01	OP-1
Reverse bits in R07	16	00	07	0000

Suppose the current value in R07 is 8009, in binary, it is:

$$(8009)_{16}=(1000\;0000\;0000\;1001)_{2}$$

All bits reversed:

$$(0111\;1111\;1111\;0110)_{2}=(7FF6)_{16}$$

Therefore the new value in the register is 7FF6.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index
OP-1	0000	Ignored

0x17 Logic Operation: Bitwise AND

Bitwise AND of two registers.

INST_REG_AND	HEX	sOP00	sOP01	sOP10	sOP11
Bit AND of R07 and R09，and store the result in R09	17	00	07	00	09

Suppose the value in R07 is 8009, and value in R09 is FF00, in binary:

$$(8009)_{16}=(1000\;0000\;0000\;1001)_{2}\\ (FF00)_{16}=(1111\;1111\;0000\;0000)_{2}$$

The AND result is:

$$(1000\;0000\;0000\;0000)_{2}=(8000)_{16}$$

8000 is stored in R07.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index
sOP10	00	Ignored
sOP11	00~3F	Register index

0x18 Logic Operation: Bitwise OR

Bitwise OR of two registers.

INST_REG_OR	HEX	sOP00	sOP01	sOP10	sOP11
Bit OR of R07 and R09，and store the result in R09	18	00	07	00	09

Suppose the value in R07 is 8009, and value in R09 is FF00, in binary:

$$(8009)_{16}=(1000\;0000\;0000\;1001)_{2}\\ (FF00)_{16}=(1111\;1111\;0000\;0000)_{2}$$

The OR result is:

$$(1111\;1111\;0000\;1001)_{2}=(FF09)_{16}$$

FF09 is stored in R07.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index
sOP10	00	Ignored
sOP11	00~3F	Register index

0x1A Arithemic Calculation: Summation

Summation of two registers.

INST_REG_SUM	HEX	sOP00	sOP01	sOP10	sOP11
Calculate the summation of values in R08 and R06，and save the result in R08	1A	00	08	00	06

Suppose R08 is 1234, R06 is 0020. The summation result 1254 is stored in R08.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index (result)
sOP10	00	Ignored
sOP11	00~3F	Register index

0x1B Arithmetic Calclation: Subtraction

Subtraction of two registers.

INST_REG_SUB	HEX	sOP00	sOP01	sOP10	sOP11
Calculate the subtraction of R08 and R06，and save the result in R08	1B	00	08	00	06

Suppose R08 is 1234, R06 is 0020. The subtraction result 1214 is stored in R08.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index (result)
sOP10	00	Ignored
sOP11	00~3F	Register index

0x1C Arithmetic Calclation: Production

Product of two registers.

INST_REG_PROD	HEX	sOP00	sOP01	sOP10	sOP11
Calculate the product of R08 and R06，and save the result in R08	1C	00	08	00	06

Suppose R08 is 1234, R06 is 0020.

$$(1234)_{16}\cdot(0020)_{16}=(2\;4680)_{16}$$

The multiplication result should be 24680, but only the least 16 bits are kept. The product 4680 is stored in R08.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index (result)
sOP10	00	Ignored
sOP11	00~3F	Register index

0x1D Arithmetic Calclation: Division (Mod)

Division of two registers.

INST_REG_DIVISION	HEX	sOP00	sOP01	sOP10	sOP11
Calculate the division of R08 and R06，and save the result in R08	1D	00	08	00	06

Suppose R08 is 1234, R06 is 0020:

$$(1234)_{16}\div(0020)_{16}=(0091)_{16}$$

The result of division is 0091, stored in R08 after the operation.

caution

Notice the division is between integers, the result is an integer qoute, instruction 1E should be excuted if remainder is needed.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index (result)
sOP10	00	Ignored
sOP11	00~3F	Register index

0x1E Arithmetic Calclation: Remainder

Calculate the division remainder of vaules in two registers.

INST_REG_REMAINDER	HEX	sOP00	sOP01	sOP10	sOP11
Calculate the remainder of division of R08 and R06，and save the result in R08	1D	00	08	00	06

Suppose R08 is 1234, R06 is 0020:

$$(1234)_{16}\%(0020)_{16}=(0014)_{16}$$

The former divided by the later results in a remainder 0014, stored in R08.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	register index (result)
sOP10	00	Ignored
sOP11	00~3F	Register index

0x1F Generate a Random Number

Generate a random 16-bit number in a target register.

INST_REG_REMAINDE	HEX	sOP00	sOP01	OP-1
Generate a random number in R05	1F	00	05	0000

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Target register index
OP-1	0000	Ignored

0x20 Byte Recomposition

Recompose the high and low bytes of two registers.

INST_BYTE_COMPOSE	HEX	sOP00	sOP01	sOP10	sOP11
Recompose the high byte of R05 and low byte of R06, and store the result in R05	20	00	05	00	06

Suppose the value in R05 is 2450, the high byte is 24; Suppose the value in R06 is FF00, the low byte is 00; The result of recomposing in R05 is 2400.

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Register index (result)
sOP10	00	Ignored
sOP11	00~3F	Register index

0x21 Byte Swap

Swap the high and low bytes of a register.

INST_BYTE_SWAP	HEX	sOP00	sOP01	OP-1
Swap the high and low bytes of R05	21	00	05	0000

info

OP/sOP	Value/Range	Remarks
sOP00	00	Ignored
sOP01	00~3F	Target register index
OP-1	0000	Ignored