Arithmetic Group of Instructions

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1. ADD
2. ADI
3. ADC
4. ACI

5. SUB
6. SUI
7. SBB
8. SBI

9. INR
10. DCR
11. INX
12. DCX

13. DAD
14. DAA

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CMP

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  Syntax: CMP R Where R can be any of the general purpose registers (A,B,C,D,E,H and L) Byte: 1byte Addressing Mode: Register   Addressing Working:  1. This is one byte instruction. 2. microprocessor will allocate one byte for opcode CMP and operands are registers i.e. operands don't need memory allocations for execution here. 3. T his instruction compares the content of the register R with content of the Accumulator. The comparison is done using subtraction of the content of register from the Accumulator .  The content of Accumulator remains unchanged . Conditions of Comarison:   1.  If [A] < [R] then Cy flag is set i.e. Cy =1 . 2. If [A] = [R]  then Z flag is set i.e Z=1. 3. If [A] > [R]  then Cy and Z flags are reset. i.e Cy = 0 and  Z = 0 4. Flag Example: CMP B where  [B] = 22H  and [A]= 11H After execution of CMP B [A] = 11H [B] = 22H   2's complement subtraction: A= 11 =  0001 0001 B= 22 =  0010 0010 1'...
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SBB

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Syntax: SBB R Where R can be any of the general purpose registers (A,B,C,D,E,H andL) Byte: 1byte Addressing Mode: Register   Addressing Working:  1.  2. A= A- R-b 3.  4. Flag Example: SBB D where  [D] = 10H , [A]= 22H and [c flag]= 1 After execution of SBB  B [A] = 11H  [D] = 10H // A= A-R-b => 22-10-1 =11H Program: write an ALP for subtraction of values present in Registers A ,B and C. Where A carries 44H and B carries 30H and C carries 10H. Store result at memory location 5000H. Label  Mnemonics  Comments Start: SUB B ;  a=a - b// 44-30= 14h            SBB  C ;  a= a - c - cy... A= 14-10-0 = 4h           STA 5000H   Stop: HLT Output: [5000H]= 04H      
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