SUB

Name	Subtract index register from accumulator with borrow
Function	Subtract a value in an index register from the accumulator, respecting the carry flag.
Syntax	SUB R
Assembled
Binary	1001 R
Decimal	144, then incrementing by 1 until 159
Hexadecimal	0x90, then incrementing by 1 until 0x9F
Symbolic
Execution	1 word, 8-bit code and an execution time of 10.8 \({\mu}\) sec
Side-effects	Depending on the result, the carry bit is reset or set.
Implemented	sub

Detailed Description

The contents of index register R are subtracted with borrow from the accumulator. The result is kept in the accumulator; the contents of R are unchanged. A borrow from the previous subtraction is indicated by the carry bit being equal to one at the beginning of this instruction. If the carry bit equals zero at the beginning of this instruction it is assumed that no borrow occurred from the previous subtraction.

This instruction sets the carry bit if there is no borrow out of the high order bit position, and resets the carry bit if there is a borrow.

The subtract with borrow operation is actually performed by complementing each bit of the contents of R and adding the resulting value plus the complement of the carry bit to the accumulator.

Notes

This instruction may be used to subtract numbers greater than 4 bits in length. The carry bit must be complemented by the program between each required subtraction operation. For an example of this, see “Decimal Subtraction”:.

The disassembly of the instruction below shows how the register is represented in the opcode:

Example programs

In order to perform a normal subtraction, the carry bit should be zero. If the accumulator contains 6, register 10 contains 2, and the carry bit is zero.

This is the set-up for the operation 6 - 2, giving the answer 4 in the accumulator.

/ Example program 1
        org    ram
        ldm    2
        xch    10
        ldm    6
        clc
        sub    10
        end

The sub operation above is carried out as follows:

   Accumulator   =   0 1 1 0
~  Register 10   =   1 1 0 1    ( register 10 = 0 0 1 0)
~  Carry         =         1    ( carry = 0)
                    ---------
    Result      1     0 1 0 0
              Carry indicates no borrow

The accumulator contains 4 and the carry bit is reset.

In this second example, if the accumulator contains 6, register 10 contains 2, and the carry bit is one:

/ Example program 2
        org    ram
        ldm    2
        xch    10
        ldm    6
        stc
        sub    10
        end

The sub operation above is carried out as follows:

   Accumulator   =   0 1 1 0
~  Register 10   =   1 1 0 1    ( register 10 = 0 0 1 0)
~  Carry         =         0    ( carry = 1)
                    ---------
    Result      1     0 0 1 1
              Carry indicates no borrow

The accumulator contains 3 and the carry bit is reset.