MCU Instructions (IV)

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1. Addition instruction without carry bit ADD A,#DATA ;例：ADD A，#10H ADD A,direct ;例：ADD A，10H ADD A,Rn ;例：ADD A，R7 ADD A,@Ri ;DO:ADD A,@R0 Purpose: Add the value in A to the value after it, and the final result is returned to A. Origin: MOV A, #30H ADD A，#10H After executing this instruction, the value in A is 40H. Practice the following questions on your own MOV 34H，#10H MOV R0，#13H MOV A，34H ADD A，R0 MOV R1，#34H ADD A，@R1

2. Addition instruction with carry bit ADDC A, Rn ADDC A, direct ADDC A,@Ri ADDC A, #data Purpose: Add the value in A to the value after it, and add the value in the carry position C. Note: Since the 51 microcontroller is an 8-bit machine, it can only perform 8-bit mathematical operations, but the range of 8-bit operations is only 0-255, which is not enough in actual work, so it is necessary to expand it. Generally, two 8-bit mathematical operations are combined into a 16-bit operation. In this way, the range of numbers that can be expressed can reach 0-65535. How to combine? In fact, it is very simple. Let's look at an example of a decimal number: 66+78。 We don't care about the process of adding these two numbers, but in fact we do it like this: first do 6+8 (low digit), then do 6+7, which is high digit. We do two additions, but we don't deliberately do it in two additions, or we don't realize that we do it twice. The reason why we do it in two additions is because these two numbers exceed the range that can be expressed by a single digit (0-9). When doing low-bit addition, a carry is generated. When we do it, we click at the appropriate position, and then add this point when doing high-bit addition. The same is true when doing 16-bit addition in the computer. Do the low 8 bits first. If the addition of two numbers generates a carry, we also need to "click" to make a mark. This mark is the carry bit C in PSW. When doing high-bit addition, this C is added. Example: 1067H+10A0H, first do 67H+A0H=107H, and 107H obviously exceeds 0FFH, so the final value stored in A is 7, and 1 goes to the CY bit in PSW. In other words, CY is equivalent to 100H. Then do 10H+10H+CY, the result is 21H, so the final result is 2107H.

3. Subtraction instruction with borrow SUBB A，Rn SUBB A,direct SUBB A,@Ri SUBB A,#data Assume (each H, (R2) = 55H, CY = 1, after executing instruction SUBB A, R2, the value in A is 73H. Note: There is no subtraction instruction without borrow. If you need to do a subtraction instruction without borrow (when doing the first subtraction), just clear CY.

4. Multiplication Instructions MUL AB The function of this instruction is to multiply the two 8-bit unsigned numbers in A and B. The result of multiplying two numbers is generally large, so the final result is expressed by a 16-bit number, with the upper 8 bits placed in B and the lower 8 bits placed in A. When the product is greater than FFFFFH (65535), OV is set to 1 (overflow), otherwise OV is 0, and CY is always 0. Example: (A) = 4EH, (B) = 5DH, execute the instruction After MUL AB, the product is 1C56H, so 1CH is placed in B and 56H is placed in A.

5. Division Instructions DIV AB The function of this instruction is to divide the 8-bit unsigned number in A by the 8-bit unsigned number in B (A/B). Division usually results in a decimal, but decimals cannot be directly expressed in computers. It uses the concept of quotient and remainder that we used when we were elementary school students and had not yet come into contact with decimals. For example, 13/5, the quotient is 2 and the remainder is 3. After the division, the quotient is placed in A and the remainder is placed in B. CY and OV are both 0. If the value in B is 00H before the division, that is, the divisor is 0, then 0V=1.

6. Add 1 instruction INC A INC Rn INC direct INC @Ri INC DPTR The purpose is very simple, that is, to add 1 to the value in the target. Example: (A) = 12H, (R0) = 33H, (21H) = 32H, (34H) = 22H, DPTR = 1234H. Execute the following instruction: INC A （A）=13H INC R2 （R0）=34H INC 21H （21H）=33H INC @R0 （34H）=23H INC DPTR （ DPTR）=1235H The results are shown above. Note: From the results, INC A and ADD A, #1 are similar, but INC A is a single-byte, single-cycle instruction, while ADD #1 is a double-byte, double-cycle instruction, and INC A will not affect the PSW bit, such as (A) = 0FFH, after INC A (A) = 00H, and CY remains unchanged. If it is ADD A, #1, then (A) = 00H, and CY must be 1. Therefore, the add 1 instruction is not suitable for addition. In fact, it is mainly used for counting, address increase, etc. In addition, addition instructions are all based on A - one of the numbers must be placed in A, and the result of the operation must also be placed in A, while the objects of the add 1 instruction are much more extensive, which can be registers, memory addresses, addresses of indirect addressing, etc.

7. Subtract 1 instruction DEC A DEC RN DEC direct DEC @Ri It is similar to the add 1 instruction, so I won’t say more. Comprehensive exercises: MOV A，#12H MOV R0，#24H MOV 21H，#56H ADD A，#12H MOV DPTR，#4316H ADD A，DPH ADD A，R0 CLR C SUBB A，DPL SUBB A，#25H INC A SETB C ADDC A, 21H INC R0 SUBB A，R0 MOV 24H，#16H CLR C ADD A，@R0 First write out the results of each step, then build the above questions and run them in software simulation, observing the changes in the contents of registers and related units to see if they are the same as your expected results.