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Arithmetic Group Instructions (Part-I)

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ADD R: 
  • This instruction adds the contents of register R with the contents of the accumulator.
  • Register R can be any general purpose register like A, B, C, D, E, H and L.
  • The result is stored in the accumulator.
  • All the flags are modified.
  • It is a one byte instruction.
  • Register addressing mode is used.
Example:
If A = 30H and B = 10H
ADD B ; This instruction adds the contents of A i.e. 30H to the contents of B i.e. 10H and the result i.e. 40H is stored in A.

ADD M:
  • This instruction adds the contents of the memory location pointed by HL register pair with the contents of the accumulator.
  • The result is stored in the accumulator.
  • All the flags are modified.
  • It is a one byte instruction.
  • Register indirect addressing mode is used.
Example:
If HL = 2000H, (2000H) = 50H and A = 10H
ADD M ; This instruction adds the contents of memory location 2000H i.e. 50H to the contents of A i.e. 10H and result 60H is stored in A.

ADC R: 
  • This instruction adds the contents of specified register to the contents of the accumulator with carry.
  • Register R can be any general purpose register like A, B, C, D, E, H and L.
  • The result is stored in the accumulator.
  • All the flags are modified.
  • It is a one byte instruction.
  • Register addressing mode is used.
Example:
If CY = 1, A = 20H and C = 30H
ADC C ; This instruction adds the contents of C (30H) to the contents of A (20H) with carry (1) and the result 51H is stored in A.

ADC M:
  • This instruction adds the contents of the memory location pointed by HL register pair with the contents of the accumulator with carry.
  • The result is stored in the accumulator.
  • All the flags are modified.
  • It is a one byte instruction.
  • Register indirect addressing mode is used.
Example:
If HL = 3000H, (3000H) = 20H, A =40H and CY = 1
ADC M ; This instruction adds the contents of memory location 3000H i.e. 20H to the contents of A i.e. 40H with carry (1) and result 61H is stored in A.
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