After instruction 2 is executed, the memory has the following values. Italics show to where data was copied and arrows show from where data was copied.

After Instruction 2 Has Been Executed

 label       address             data                  comment

         0000000000000000  0000000001000000 <----|start0000000000001000
         0000000000000001  0000000000001001      | number to increment
         0000000000000010  0000000000010000 <--| | answer
                                               | |
instr_1  0000000000001000  0000000000000001    | | from address 
         0000000000001001  0000000000001100    | | to address
         0000000000001010  0000000000001111    | | 'to' bits to copy to
         0000000000001011  0000000000110000    | | instr.addr.and rot.
instr_2  0000000000001100  0000000001001001    | | from address 
         0000000000001101  0000000000000010    | | to address
         0000000000001110  0000000011111111    | | 'to' bits to copy to
         0000000000001111  0000000001000000 ---+-| instr.addr.and rot.
instr_3  0000000000010000  0000000000000000    |   from address 
         0000000000010001  0000000000000000    |   to address
         0000000000010010  0000000000000000    |   'to' bits to copy to
         0000000000010011  0000000001000000    |   instr.addr.and rot.
                                               |
add1tabl 0000000001000000  0000000000000001    |
         0000000001000001  0000000000000010    |
         0000000001000010  0000000000000011    |
         0000000001000011  0000000000000100    |
         0000000001000100  0000000000000101    |
         0000000001000101  0000000000000110    |
         0000000001000110  0000000000000111    |
         0000000001000111  0000000000001000    |
         0000000001001000  0000000000001001    |
         0000000001001001  0000000000010000 ---|

Now, the result of 1 being added to 9 is 10 and the result, 10, written as 00010000, is stored in the last eight bits of latch 0000000000000010.

Instruction 3 ('instr_3') does nothing but execute over and over until the processor is stopped.

The program calculates what one more than nine is and finds that the answer is ten. Nine is represented as 1001 in latch 0000000000000001 and ten is represented as 00010000 in latch 0000000000000010.

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