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3 DEFM " ADD HL,HL
2 the number
2 LD HL,STORE
2 LD HL,22528
2 EXAMPLE OF RELATIVE JUMPS
2 0 0 0 0 0 1 0 1 +5
2 LD HL,STORE
2 LD (HL),A
2 DJNZ LOOP
1 transfer (HL) to (DE), increment HL and DE, decrement BC, set PV flag if BC=0
1 transfer (HL) to (DE), decrement HL, DE, and BC, set PV flag if BC=0
1 the port whose number is in C
1 the memory location (IY)
1 the memory location (IX)
1 the memory location (DE)
1 the memory location (BC)
1 the location (HL)
1 the contents of the Accumulator is eXclusively ORed with
1 the contents of
1 the computer is HALTed until an interrupt comes to awake it again
1 the carry flag is set, ie Set Carry Flag
1 the carry flag is inverted, ie Compliment Carry Flag
1 the alternative AF registers
1 the accumulator is ComPared with
1 the Stack Pointer
1 the Refresh register
1 the RETurn address is POPped from the stack
1 the L register
1 the Interrupt vector register
1 the IY register
1 the IX register
1 the HL register
1 the H register
1 the Flag register
1 the E register
1 the DE register
1 the D register
1 the C register
1 the BC, DE, HL registers are all EXchanged with the alternative set
1 the BC register
1 the B register
1 the Accumulator is logically ORed with
1 the Accumulator is logically ANDed with
1 the Accumulator is NEGated, ie made negative
1 the Accumulator
1 the AF registers
1 search from (HL) Incrementing BC bytes for data in Accumulator
1 search from (HL) Decrementing BC bytes for data in Accumulator
1 rotate the accumulator and carry Right
1 rotate the Accumulator and carry Left
1 rotate the Accumulator Right, shifting bit 0 into Carry
1 rotate the Accumulator Left shifting bit 7 into Carry
1 rotate Right Decimal contents of (HL) with least significant end of Accumulator
1 rotate Left Decimal contents of (HL) with least significant end of Accumulator
1 reSTart at address
1 push the return address to stack and CALL
1 port with the above number
1 perform relative jump to
1 perform jump to
1 output from (HL) to port (C) increment HL, decrement B
1 ouTput to port (C), from block from (HL) Decrementing HL for B bytes
1 ouTput from (HL) to port (C) block from (HL) Incrementing for B bytes
1 oUTput from (HL) to port (C) Decrement HL, decrement B
1 memory location
1 location (IY+offset)
1 location (IX+offset)
1 is loaded with
1 is Shifted Right Logically, 0 is shifted into bit 7
1 is Shifted Right Arithmetically, sign bit unchanged
1 is Shifted Left Arithmetically, 0 is shifted into bit 0
1 is Rotated Right, shifting bit 0 into Carry
1 is Rotated Left shifting bit 7 into Carry
1 is PUSHed onto the stack and the SP decremented by 2
1 is POPped from the stack and the SP incremented by 2
1 is INCremented by 1
1 is EXchanged with
1 is DECremented by 1
1 invert all bits in A register, ie ComPLiment
1 input data from
1 if the Zero flag is set,
1 if the Zero flag is not set,
1 if the Sign flag is set, ie Minus,
1 if the Sign flag is not set, ie Positive,
1 if the Parity flag is sEt, ie Parity is Even,
1 if the Parity flag is nOt set, ie Parity is Odd,
1 if the Carry flag is set,
1 if the Carry flag is not set,
1 iNput into (HL) from port (C) increment HL, decrement B
1 iNput into (HL) from port (C) Decrement HL, decrement B
1 iNput from port (C), into block from (HL) Incrementing HL for B bytes
1 iNput from port (C), into block from (HL) Decrementing HL for B bytes
1 i.e. carry ' ' ' '
1 i.e. borrow 0 1 1 2 0 2
1 i.e. B8 hex = (B hex)* 16+ (8)* 1
1 has taken from it the carry and
1 has added to it the carry and
1 has added to it
1 enable Interrupts. NOT PERFORMED BY SIMULATOR.
1 disable Interrupts. NOT PERFORMED BY SIMULATOR.
1 decrement B, if zero continue, if Not Zero perform relative jump to
1 decimal Adjust Accumulator
1 compare (HL) with A, increment HL, decrement BC
1 compare (HL) with A, decrement HL, and BC
1 change Interrupt Mode to
1 block transfer (BC) bytes starting at (HL) to (DE) incrementing
1 block transfer (BC) bytes starting at (HL) to (DE) decrementing
1 and Carry are rotated Left
1 and Carry are Rotated Right
1 This is the number we originally assumedat the beginning of our discussion.
1 The Accumulator has subtracted from it
1 Space missing
1 Since the contents of the HL registerpair can be the result of some calculationthe jump can be conditional.
1 RETurn from Non-maskable interrupt: NOT PERFORMED BY SIMULATOR
1 RETurn from Interrupt: NOT PERFORMED BY SIMULATOR
1 One difficulty with the above types ofjump is that, if the start address of theprogram were changed, all the jumpaddresses would require modification.Relative jumps do not have thisdifficulty. The absolute address is notstored with the instruction. It is thedifference between the current contents ofthe program counter and the address towhich we wish to jump that is storedwithin the instruction. Relative jumps arelimited to jumping backwards 128 locationsand forwards 127 locations.
1 Offset too big
1 Offset missing
1 Number too large
1 Number missing
1 No OPeration is performed, inspite of this it takes .
1 No DEFB, or DEFW
1 NOT DEFINED
1 Missing space or ,
1 Label too long
1 Label not found
1 JP Z,nn jump on zero
1 JP NZ,nn jump on not zero
1 JP NZ,nn jump if zero flag not set
1 JP NC,nn jump on no carry
1 JP NC,nn jump if carry flag not set
1 JP C,nn jump on carry
1 Instruction unknown
1 Hence to make a positive number negativewe invert all its binary bits, known asComplimenting, and add 1. The notation isgenerally termed 2s complement
1 FLAGS$ Registers and Contents % STAC
1 Error after inst.
1 EXAMPLES OF BINARY CODED DECIMAL NOTATION
1 EX5B:DEFM " LD HL,(STORE)
1 EFM " INC DE
1 DJNZ e decrement and jump on non zero
1 DEFM "using addition to successively multiply by"
1 DEFM "number held in the HL register pair by 10"
1 DEFM " LD HL,(STORE)
1 DEFM " LD DE,(STORE)
1 DEFM " LD (STORE),HL
1 DEFM " JR LOOP1
1 DEFM " INC HL
1 DEFM " INC H
1 DEFM " DJNZ LOOP2
1 DEFM " ADD HL,DE
1 Cursor keys, delete, and auto-repeat are as standard. BREAK returns you to menu.
1 Can only ADD IX/IY
1 Bracket missing
1 ADDRESS MACHINE CODE
1 101, which is 1 less thanthat for -2.
1 10. Bit 7, the most significantbit, represents the sign ( + or - ) ofthe number. When bit 7 is "1" the numberis negative, and when "0" the number ispositive.
1 10 for an example.Inverting all bits of the binary number of+2 gives
1 1, or the processor would notbe able to take 6 from 5 and get theright answer. Further 5 - 7 = -2 or 254or
1 1 or 127 and the largestnegative number 1
1 010 and its negativeequivalent -2 or
1 01 seconds to do it
1 0 or -128.
1 -^1^>^E^V^O^]^j^q^{^
1 where e is the displacement inJR NC,e the range 127 to -128
1 the standardcalculation.
1 since each digit can have one of tenstates, designated 0 - 9.
1 operation.
1 of the Accumulator.
1 jump if zero flag set
1 jump if carry flag set
1 i.e. -2 is equivalent to 254 etc.
1 ectively multiplies a"
1 contents of the Accumulator by
1 complementing and adding 1 in one
1 bit 7 6 5 4 3 2 1 0
1 as BC, DE, HL,or SP.
1 You are trying to run code in allocated storage area. RET missing? Please EDIT
1 You are about to write to memory which will affect your program. Please edit it.
1 You are about to affect memory area not allocated to you. Please edit program.
1 When performing arithmetic operationswhich we wish to interpret within therange -128 to +127, the Carry flag nolonger signals an out of range result.However another flag, the overflow (P/V)flag does. It is a "1" whenever the answeris outside the range -128 to +127 and "0"within the range. It is effectively acarry into bit 7 which would make the signbit incorrect.
1 We would still like to be able to performcalculations on decimal numbers withouthaving to convert to binary or hexadecimalform. Hexidecimal form holds a key to amethod of achieving this.
1 We shall then look at numbers held inregisters in more detail.
1 We require 16 different symbols torepresent a digit. 0-9 gives us thefirst 10 and A-F the final 6. A listof decimal, binary, and hexadecimalequivalents are given on the next screen.
1 To perform this operation the Z80 usestwo further flags, the H flag ( a Halfcarry between nibbles ), and N flag( subtract instruction performed last ).Since these flags cannot be tested as partof a conditional jump etc. they are oflittle use.
1 To get round this problem use a label,declaring it in the label column in frontof the instruction to which you wish tojump.
1 Thisinstruction is used after an 8 bit ADD,ADC, SUB, or SBC to adjust the Accumulatorcontents by adding 6 to the nibble, ifeither nibble in the answer is above 9.
1 This simple example counts up the numberof memory locations holding a number above192 and below 64 in the first 32 memorylocations.
1 This presentation is termed Binary CodedDecimal or BCD.
1 This example illustrates a method ofmultiplication. There are other fastermethods that will be shown later.
1 They are simply those in which DE, or BCcan take part. The instruction equivalentto EX DE,HL is however EX (SP),HL, thatEXchanges the contents of the bottom ofthe Stack with the HL register.
1 These examples show simple addition andsubtraction in BCD format. Remember tokeep in Hex display mode or theillustrations will not appear in BCD.
1 There is no reason for sticking to 8 bitsusing this method. So long as the numberof bits is sufficient and the mostsignificant bit is taken as the sign bitany size positive or negative number canbe represented.
1 There are too many registers called up, to display. Program will RUN as normal.
1 There is one particular relative jumpinstruction that is particularly usefulnamely Decrement and Jump on Non-Zero, orDJNZ. The B register is decremented and,if the result is not zero, the processorperforms a relative jump to the computednew address. If the result is zero thenthe next instruction in sequence is taken.DJNZ allows us to repeat a series ofinstructions a preset number of times. Thenumber of repeats is equal to the contentsof the B register when the loop isentered, so long as the B register doesnot form part of the series ofinstructions.
1 The stack pointer is outside allocated memory area. Please edit program.
1 The simulator uses the memory area 32000to 32255 for the pupils programs. TheStack Pointer is initially set to 32254,and works its way down towards the programarea. Keep the SP at the top end of thisarea of memory to ensure that thesimulator does not stop.
1 The simplest jump is JP nn, where nn isthe address to which the program is tojump. This is an unconditional jump, theProgram Counter always jumping to theaddress nn.
1 The program has been completed. You may EDIT or reRUN. Exit by pressing BREAK.
1 The mnemonic for relative jumps is JR andcan take all the forms discussed so farfor JP, i.e. JR , JR NC, JR C, JR NZ andJR Z.
1 The maximum number that can be held in an8 bit register is
1 The last four pairs of memory locationsin the Stack are displayed at the bottomright hand corner of the screen. As withthe Program Counter, the position to whichthe Stack Pointer points is shown in Cyan.
1 The instruction is PUSH dd. Continuing toPUSH data on to the Stack results in thedata being pushed in at the bottom of theStack and the Stack Pointer working itsway down through memory.
1 The instruction INC SP is included at theend of the program to ensure that SP isdisplayed. The actual value in the SPregister is immaterial for most of itsapplications.
1 The assembler, within this tutor, has oneinstruction not normally found in anassembler, namely BIN. The assemblerallocates one memory location to the line,which would normally be labelled. Itconstantly displays this location inbinary form. The number after BIN is thenumber initially inserted into thelocation.
1 The Spectrum works on 32 characters perline and not 42 as does the TUTOR
1 The BC, DE and HL registers may be PUSHedon to and POPped from the Stack. TheAccumulator may also be PUSHed on to theStack, but this is done with the flagregister F, which will be discussed later.
1 The return instruction RET, POPs thereturn address from the Stack, back intothe PC register, to continue where themain program left off. If the number ofPUSHes and CALLs in the subroutine doesnot equal the number of POPs and RETurns,obviously this will not happen.
1 The processor, under the programmer'sdirection, has the capacity to makedecisions. In the following lessons weshall discuss these instructions that makea computer more than just a calculator.
1 The location 23060 holds a number 256less than the start of the Spectrumcharacter set. This example writescharacters along the top line of thescreen. Once you have followed theindividual steps of the program, keepinga key pressed will result in the programbeing stepped through quickly.
1 The largest positive number we can holdin a single register, using this notationis 0
1 The hex digit requires adjustment toavoid the digits A-F. Adding 6 to thedigit, if these symbols appear, results incorrect adjustment.
1 The aspect of a computer that makes itmore than a calculator is its ability tochange its sequence of operations as aresult of earlier instructions. Withinthe Z80 there is a counter called theProgram Counter or PC. This counter keepstrack of which instruction the processoris to perform next. By modifying itscontents, the processor can be jumped toany instruction in memory.
1 The arrangement of the screen, and otheruseful memory locations within theSpectrum are given in the TUTOR manual.
1 The Z80 has two instructions that may beused for these operations
1 The Stack is an area of external RandomAccess Memory, that is used as a series ofregister pairs, in conjunction with aStack Pointer. The Stack Pointer (SP) is asingle double length register within theZ80, capable of holding up to 65335.
1 The Stack Pointer can take part in anumber of instructions already discussed.
1 The Simulator can be switched betweendecimal and hexadecimal display bypressing "g", when a program is notrunning.
1 The Sign and overflow flags are alsooperative after ADC and SBC instruction onthe HL register. It reflects the 15th bit(or bit 7 of H register) .
1 The operation luckily also makes anegative number positive. A veryimportant point.
1 The first example shows nesting ofsubroutines : a subroutine being calledfrom within another subroutine.
1 The conditional jumps involving thecarry and zero flags are:-
1 The Z80 allows this feature by theinstruction CALL. The CALL instruction issimilar to the JP instruction except thatthe address of the next sequentialinstruction, held in the PC register, isPUSHed onto the Stack, prior to the jump.
1 The Z80 accommodates the aboveadjustment, with the instruction DAA,Decimal Adjust Accumulator.
1 The more powerful jump is theconditional jump. The processor tests thestate of a flag, before either jumping orcontinuing in sequence.
1 Taking our number 10111000 for exampleand noting that bits that are "0"contribute nothing to the number, thedecimal equivalent of 10111000 is given by
1 Take the hex number C (= 12 dec.). Adding6 gives 18 decimal or 12 hex. Hence thehex number looks identical to the decimalanswer required.
1 THE STACK
1 Summary - the Stack and SP register
1 Summary - relative jumps
1 Summary - jumps
1 Summary - calls and returns
1 So far we have only dealt with positivenumbers. If we had taken 6 from 5 we wouldend up with the answer 255 with the carryset. There is an interpretation of binarynumbers that allows us to consider thisanswer as the negative number -1.
1 Similar we can perform subtraction.
1 STORE DEFW 0
1 STORE DEFB 64
1 STORE DEFB 6
1 STORE DEFB 124
1 STORE BIN 15
1 SR2 LD A,(HL)
1 SR1 CALL SR2
1 SOURCE CODE
1 SBC HL,SP
1 Return to operating system
1 RET NZ Not Zero
1 RET NC No Carry
1 RELATIVE JUMPS
1 Program stopped. You may re-RUN, EDIT or press BREAK to return to menu.
1 Program has successfully assembled. Press RUN, EDIT, or exit by pressing BREAK.
1 Program counter has jumped to non-valid address. EDIT or reRUN program.
1 Press SPACE to select and ENTER to start
1 Press SPACE to continue.
1 Press RUN to begin example, EDIT to edit, and BREAK to return to menu.
1 Please confirm disk loading required Y/N
1 Performing a similar calculation to theabove we can show that a register pair canrepresent a number in the range +32767 to-32768.
1 PUSH DD where DD is AF, BC, DE, or HL.
1 POP dd retrieves data from the Stack intothe defined register pair. The registerpair is loaded with the contents of thememory location pointed to by SP. Then theSP register is incremented twice.
1 Now that hex notation has been introducedwe can discuss how the computer storesprograms in more detail. As alreadymentioned, the instructions are stored inmemory as numbers. Some instructions onlyrequire a single number, others requiremore. Column 2 on the Simulator displayshows the memory contents for theinstructions in hex form.
1 No flags are affected by any jumpinstructions including DJNZ.
1 NUM2 BIN 95
1 NUM1 BIN 81
1 NEG negates, or makes negative, the
1 Luckily assemblers takes account of allthese complexities, and you will not berequired to learn the tables inAppendix A of your Spectrum Manual.
1 Loading lessons 26-35
1 Loading lessons 18-25
1 Loading lessons 10-17
1 Loading lessons 1-9
1 Loading has been stopped
1 Loading error
1 Load further lessons from disk
1 Lessons 10 - 17
1 LOOP2 LD A,(DE)
1 LOOP2 DEC (HL)
1 LOOP1 LD (STORE),HL
1 LOOP1 LD (HL),A
1 LOOP LD A,(HL)
1 LOOP DEC (HL)
1 LOOP ADD HL,DE
1 LD SP,(nn)
1 LD HL,56789
1 LD DE,(23606)
1 LD (nn),SP
1 Just as we add two decimal numberstogether, carrying one forward if the sumof two digits is greater than 9, we canadd two binary numbers. If the sum isabove one, a one is carried forward.
1 It must be remembered that a subroutinemay require the use of a register whosecontents are still required in the mainprogram. Its contents must therefore bePUSHed onto the stack or otherwise storedelsewhere, and retrieved at the end of thesubroutine.
1 It is important to remember that the ZeroCarry, Sign, and overflow flags are alwaysoperative after an 8 bit arithmeticinstruction or a 16 bit addition involvingthe Carry. Your interpretation of theresult determines which flag(s) you shouldbe interested in.
1 Indirect jumps are also available on theZ80 using the HL register pair. Theinstruction JP (HL) loads the programcounter with the contents of the HLregister. Thus the address of the nextinstruction to be performed is equal to the contents of the HL register.
1 In this interpretation -1 must equal 255or
1 In a register each digit or bit has onlytwo states. Therefore we can interpret thedigits as having weightings of
1 If we were to take the top off a Z80 chipand, using a powerful microscope, look ata register, we would see that it consistsof 8 cells. These cells are capable ofbeing switched between two differentstates. Conventionally we designate onestate as "1" and the other as "0". Thepattern of 1's and 0's could be 10111000say.
1 If the condition is not met, the programwill not CALL the subroutine or RETurnfrom it.
1 If a series of registers is PUSHed on tothe Stack, and subsequently POPped off theStack, they reappear in the reverse order.
1 INTRODUCTION
1 How do we interpret this pattern? Theamswer is, anyway we like. In this lessonwe show that this pattern could be184, -72, B8H, and even CP B. The only onethe processor has built into it is thelatter.
1 HEXADECIMAL NOTATION
1 From now on dd can be considered
1 Ex - unconditional calls & rets
1 Ex - relative jumps 2
1 Ex - relative jumps 1
1 Ex - pushes and pops
1 Ex - jumps
1 Ex - hex notation
1 Ex - conditional calls
1 Ex - binary notation
1 Ex - BCD notation
1 Ex - +ve and -ve numbers
1 Each digit in a number can be consideredto have a weighting. For decimal numbersthese weightings are
1 Each digit is held in 4 bits of theregister. These 4 bits are called anibble, the whole 8 bit word is termed abyte. Who said programmers have no senseof humour?
1 EXAMPLES USING THE STACK
1 EXAMPLES OF JUMPS
1 EXAMPLES OF HEXADECIMAL NOTATION
1 EXAMPLES OF BINARY NOTATION
1 EXAMPLES OF +VE AND -VE NOTATION
1 EXAMPLE OF CONDITIONAL CALLS
1 EXAMPLE OF CALLS TO SUBROUTINE
1 EX (SP),HL
1 Disk V1.0 (c) HanSoft '85
1 Conversion of a hex number to its decimalequivalent
1 Continuing to POP data from the Stackresults in information being POPped fromthe bottom of the Stack and the StackPointer working its way back up throughthe memory.
1 Carry set
1 CPL complements or inverts the contents
1 CALLS TO SUBROUTINES
1 Both conditional and unconditional CALLsand RETurns are available on the Z80. Allfour types of conditions alreadyconsidered ( NC, C, NZ and Z ) can beused with CALLs and RETurns.
1 Binary numbers are too cumbersome anddecimal too inconvenient to displaymultiple register numbers. A method ofpresenting binary numbers has beendeveloped to get round these problems.Instead of basing the number on 10 as indecimal or 2 as in binary, we base it on16. This numbering system is calledHexadecimal.
1 Before loading the Stack from anyinternal register pair, the Stack Pointeris decremented twice, ( thus pointing tothe next pair of memory locations ). Thecontents of the register is then loadedinto the Stack indirectly using the SPregister as the address.
1 BINARY NOTATION
1 BINARY CODED DECIMAL NOTATION
1 BELOW INC D
1 Always remember that a hex number has abase of 16 decimal, a decimal number has abase of 10 and a binary number a base of2.
1 Although you can enter the address withinthe jump in number form, if it does notequal the address of the start of aninstruction, the simulator will stop atthe JUMP. This is because the processorwould perform the instruction it findsat that address and anything could happen,and usually does.
1 Although these examples appear to be allpositive, they can be viewed equally asnegative where appropriate and the Signand overflow flags observed operating.Remember a number above 128 is a negativenumber. Since all the negative numbers inthe examples are small, they can be seenquickly and simply be converted bysubtracting the number from 256.
1 All assemblers calculate the displacementfor relative jumps. The assembler withinthis TUTOR is no exception. Relative jumpsshould be entered with labels.
1 All the numbers used to describe aninstruction are given in sequence alongthe line. Notice that numbers called up inan instruction appear towards the end ofthe string of numbers. The earlier numbersdetermine the type of instruction. Sincethe total number of instructions exceed256, a prefix number is used to produceother sets of instructions. The prefixesyou will find are EDH, CBH, DDH and FDH.
1 All 16 states of the 4 binary bits arecovered by 0-F. Hence only 2 hexadecimaldigits are required to display thecontents of an 8 bit register. Our example10111000 divides up into 1011 1000 orB8 Hex. A double register holding say11001001 10111000 will be represented byC9B8 hex.
1 ANS DEFW 0
1 ANS BIN 0
1 ADD HL,SP
1 ADC HL,SP
1 ABOVE INC E
1 A subroutine is a part of a program thatis called from the main program and oncecompleted returns back to the mainprogram. The subroutine is usually calledfrom a number of places in the program.
1 A Sign flag (S) is provided on the Z80.It duplicates the sign ( bit 7 ) of theanswer after any arithmetic operation onthe Accumulator. The sign flag is "0" fora positive result and "1" for a negativeresult.
1 A useful operation would be to make apositive number negative. Changing bit 7does not do this. Take the positivenumber 2 or
1 = 11*16+8=184
1 3 2 1
1 17. Positive and negative notation
1 16. Binary Coded Decimal notation
1 15. Hexadecimal notation
1 14. Binary notation
1 13. Calls to subroutines
1 128+64+32+16+8+4+2+1=255
1 128+32+16+8=184.
1 128 64 32 16 8 4 2 1
1 12. The Stack
1 11. Relative jumps
1 1000 100 10 1
1 10. Conditional and unconditional jumps
1 1 1 1 1 1 0 1 1 -5
1 1 0 0 1 0 0 1 1
1 0 1 1 0 1 1 1 0
1 0 1 1 0 0 1 1 0
1 0 0 1 1 1 0 0 1
1 - 0 0 1 0 1 1 0 1
1 + 0 0 1 0 0 1 0 1
1 & clears all ENTER tabs to next line STOP assembles SPACE tabs in label are
1 Loading error
1 LD HL,NUM2
1 LD HL,4589H
1 LD HL,16384
1 LD HL,(STORE)
1 LD E,(HL)
1 LD DE,FFFEH
1 LD DE,3812H
1 LD DE,34567
1 LD DE,2468H
1 LD BC,12345
1 LD B,(HL)
1 LD A,(NUM1)
1 LD (ANS),HL
1 LD (ANS),A
1 JR NZ,LOOP
1 JR LOOP1
1 JP NZ,LOOP2
1 JP NZ,LOOP1
1 INC (HL)
1 Error found during assembly. Please correct before running program.
1 EX (SP),HL
1 DJNZ SR1
1 DJNZ LOOP2
1 CALL SR1
1 CALL NC,ABOVE
1 CALL C,BELOW
1 ADD HL,DE
1 ADD A,128
1 ADD A,(HL)
1 CALL Z,nn
1 CALL NZ,nn
1 CALL NC,nn
1 CALL C,nn
1 ASSEMBLED CODE LABELS INSTRUCTIONS
1 invert 1 1 1 1 1 0 1 0
1 invert 0 0 0 0 0 1 0 0
1 POSITIVE AND NEGATIVE NUMBER NOTATION
1 Enter the Z-80 simulator ? Y/N
1 Decimal Binary Hexadecimal