# Learn 6502 assembly

The 6502 handles data in its registers, each of which holds one byte
(8-bits) of data.

There are three general use and two special
purpose registers:

accumulator (A) – Handles all arithmetic and logic

X and Y – General purpose registers, with limited abilities

S – Stack pointer

P – Processor status. Holds the result of tests
and flags

Stack Pointer

instruction it needs to know where to return when finished. The 6502
keeps this information in low memory, from \$0100 to \$01FF and uses the
stack pointer as an offset. The stack grows down from \$01FF and makes
it possible to nest subroutines up to 128 levels deep.

The processor status register is not directly accessible by any 6502
instruction. Instead, there are instructions that test the
bits of the processor status register.

The flags within the register are:

bit -> 7 0
+—+—+—+—+—+—+—+—+
| N | V | | B | D | I | Z | C | <– flag, 0/1 = reset/set
+—+—+—+—+—+—+—+—+

N = NEGATIVE. Set if bit 7 of the accumulator is set.

``````   V  =  OVERFLOW. Set if the addition of two like-signed numbers or the
subtraction of two unlike-signed numbers produces a result
greater than +127 or less than -128.

B  =  BRK COMMAND. Set if an interrupt caused by a BRK, reset if
caused by an external interrupt.

D  =  DECIMAL MODE. Set if decimal mode active.

I  =  IRQ DISABLE.  Set if maskable interrupts are disabled.

Z  =  ZERO.  Set if the result of the last operation (load/inc/dec/

C  =  CARRY. Set if the add produced a carry, or if the subtraction
produced a borrow.  Also holds bits after a logical shift.
``````

Accumulator

The majority of the 6502’s business makes use of the accumulator. All
addition and subtraction is done in the accumulator. It also handles
the majority of the logical comparisons (is A > B ?) and logical bit
shifts.

X and Y

These are index registers often used to hold offsets to memory
locations. They can also be used for holding needed values. Much of
their use lies in supporting some of the addressing modes.

The 6502 has 13 addressing modes (ways of accessing memory). The 65C02

They are:

``````  +---------------------+--------------------------+
|      mode           |     assembler format     |
+=====================+==========================+
| Immediate           |          #aa             |
| Absolute            |          aaaa            |
| Zero Page           |          aa              |   Note:
| Implied             |                          |
| Indirect Absolute   |          (aaaa)          |     aa = 2 hex digits
| Absolute Indexed,X  |          aaaa,X          |          as \$FF
| Absolute Indexed,Y  |          aaaa,Y          |
| Zero Page Indexed,X |          aa,X            |     aaaa = 4 hex
| Zero Page Indexed,Y |          aa,Y            |          digits as
| Indexed Indirect    |          (aa,X)          |          \$FFFF
| Indirect Indexed    |          (aa),Y          |
| Relative            |          aaaa            |     Can also be
| Accumulator         |          A               |     assembler labels
+---------------------+--------------------------+``````

The value given is a number to be used immediately by the
instruction. For example, LDA #\$99 loads the value \$99 into the
accumulator.

The value given is the address (16-bits) of a memory location that
contains the 8-bit value to be used. For example, STA \$3E32 stores
the present value of the accumulator in memory location \$3E32.

The first 256 memory locations (\$0000-00FF) are called “zero page”. The
next 256 instructions (\$0100-01FF) are page 1, etc. Instructions
making use of the zero page saves memory, by not using an extra \$00 to
indicate the high part of the address.

For example,

LDA \$0023 — works but uses an extra byte
LDA \$23 — the zero page address

Many instructions are only one byte in length and do not reference
memory. These are said to be using implied addressing.

For example,

``````  CLC  -- Clear the carry flag
DEX  -- Decrement the X register by one
TYA  -- Transfer the Y register to the accumulator``````

Only used by JMP (JuMP). It takes the given address and uses it as a
pointer to the low part of a 16-bit address in memory, then jumps to

For example,

``````  JMP (\$2345)   -- jump to the address in \$2345 low and \$2346 high

So if \$2345 contains \$EA and \$2346 contains \$12 then the next
instruction executed is the one stored at \$12EA.  Remember, the
6502 puts its addresses in low/high format.``````

The final address is found by taking the given address as a base and
adding the current value of the X or Y register to it as an offset. So,

``  LDA \$F453,X  where X contains 3``

Load the accumulator with the contents of address \$F453 + 3 = \$F456.

Same as Absolute Indexed but the given address is in the zero page
thereby saving a byte of memory.

Find the 16-bit address starting at the given location plus the
current X register. The value is the contents of that address.

For example,

``  LDA (\$B4,X)  where X contains 6``

This gives an address of \$B4 + 6 = \$BA. If \$BA and \$BB contain \$12 and
\$EE respectively, then the final address is \$EE12. The value at
location \$EE12 is put in the accumulator.

Find the 16-bit address contained in the given location ( and the one
Fetch the value stored at that address. For example,

``  LDA (\$B4),Y  where Y contains 6``

If \$B4 contains \$EE and \$B5 contains \$12 then the value at memory
location \$12EE + Y (6) = \$12F4 is fetched and put in the accumulator.

The 6502 branch instructions use relative addressing. The next byte
is a signed offset from the current address, and the net sum is the
address of the next instruction executed.

For example,

``  BNE \$7F   (branch on zero flag reset)``

will add 127 to the current program counter (address to execute) and
start executing the instruction at that address.

SImilarly,

``  BEQ \$F9   (branch on zero flag set)``

will add a -7 to the current program counter and start execution at

Remember, if one treats the highest bit (bit 7) of a byte as a sign (0
= positive, 1 = negative) then it is possible to have numbers in the
range -128 (\$80) to +127 (7F). So, if the high bit is set, i.e. the
number is > \$7F, it is a negative branch. How far is the branch? If
the value is < \$80 (positive) it is simply that many bytes. If the value is > \$7F (negative) then it is the 2’s compliment of the given
value in the negative direction.

``````  2's compilment
--------------

The 2's compilment of a number is found by switching all the bits
from 0 -> 1 and 1 -> 0, then adding 1.  So,

\$FF  =  1111 1111   <-- original
0000 0000   <-- 1's compliment
+          1
---------
0000 0001   <-- 2's compliment, therefore \$FF = -1

Note that QForth uses this for numbers greater than 32768 so that
65535 = -1 and 32768 = -32768.``````

In practice, the assembly language programmer uses a label and the
assembler takes care of the actual computation. Note that branches
can only be to addresses within -128 to +127 bytes from the present

Like implied addressing, the object of the instruction is the
accumulator and need not be specified.

## The 6502 Instruction Set

There are 56 instructions in the 6502, and more in the 65C02. Many
instructions make use of more than one addressing mode and each
opcode that specifies it exactly.

So,

``````  A9  =  LDA #\$aa   Immediate addressing mode load of accumulator
etc.``````

Some 6502 instructions make use of bitwise logic. This includes AND,
OR, and EOR (Exclusive-OR). The tables below illustrate the effects
of these operations:

``````  AND   1  1  ->  1    "both"
1  0  ->  0
0  1  ->  0
0  0  ->  0

OR    1  1  ->  1    "either one or both"
1  0  ->  1
0  1  ->  1
0  0  ->  0

EOR   1  1  ->  0    "one or the other but not both"
1  0  ->  1
0  1  ->  1
0  0  ->  0

Therefore,  \$FF AND \$0F  =  \$0F since,

1111 1111
and  0000 1111
---------
0000 1111  = \$0F``````

AND is useful for masking bits. For example, to mask the high order
bits of a value AND with \$0F:

``  \$36 AND \$0F  =  \$06``

OR is useful for setting a particular bit:

``````  \$80 OR \$08   =  \$88

since  1000 0000  (\$80)
0000 1000  (\$08)
or ---------
1000 1000  (\$88)``````

EOR is useful for flipping bits:

``````  \$AA EOR \$FF  =  \$55

since  1010 1010  (\$AA)
1111 1111  (\$FF)
eor ---------
0101 0101  (\$55)``````

Other 6502 instructions shift bits to the right or the left or rotate
them right or left. Note that shifting to the left by one bit is the
same as multipling by 2 and that shifting right by one bit is the same
as dividing by 2.

The 6502 instructions fall into 10 groups, with two odd
instructions – NOP and BRK:

``````  Load and Store Instructions
Arithmetic Instructions
Increment and Decrement Instructions
Logical Instructions
Jump, Branch, Compare and Test Bits Instructions
Shift and Rotate Instructions
Transfer Instructions
Stack Instructions
Subroutine Instructions
Set/Reset Instructions
NOP/BRK Instructions``````

LDX – LoaD the X register
LDY – LoaD the Y register

STA – STore the Accumulator
STX – STore the X register
STY – STore the Y register

Microprocessors spend much of their time moving stuff around in
memory. Data from one location is loaded into a register and stored
in another location, often with something added or subtracted in the
process. Memory can be loaded directly into the A, X, and Y registers
but as usual, the accumulator has more addressing modes available.

If the high bit (left most, bit 7) is set when loaded the N flag on
the processor status register is set. If the loaded value is zero the
Z flag is set.

Arithmetic Instructions

SBC – SuBtract from accumulator with Carry

The 6502 has two arithmetic modes, binary and decimal. Both addition
and subtraction implement the carry flag to track carries and borrows. Note that in the case of
subtraction it is necessary to SET the carry flag as it is the opposite of the carry that is subtracted.

CLC
.
.
.
.
.

Clear the carry flag, and perform all the additions. The carry
between additions will be handled in the carry flag. Add from low
byte to high byte. Symbolically, the net effect of an ADC instruction is:

A + M + C –> A

Subtraction follows the same format:

SEC
SBC …
.
.
SBC …
.
.
.

In this case set the carry flag first and then do the subtractions.
Symbolically,

A – M – ~C –> A , where ~C is the opposite of C

Ex.1

``````    A 16-bit addition routine.  \$20,\$21 + \$22,\$23 = \$24,\$25

CLC         clear the carry
LDA \$20     get the low byte of the first number
STA \$24     store in the low byte of the result
LDA \$21     get the high byte of the first number
ADC \$23     add to it the high byte of the second, plus carry
STA \$25     store in high byte of the result

... on exit the carry will be set if the result could not be
contained in 16-bit number.``````
``````    A 16-bit subtraction routine.  \$20,\$21 - \$22,\$23 = \$24,\$25

SEC         clear the carry
LDA \$20     get the low byte of the first number
SBC \$22     add to it the low byte of the second
STA \$24     store in the low byte of the result
LDA \$21     get the high byte of the first number
SBC \$23     add to it the high byte of the second, plus carry
STA \$25     store in high byte of the result

... on exit the carry will be set if the result produced a
borrow``````

Aside from the carry flag, arithmetic instructions also affect the N,
Z, and V flags as follows:

``````  Z = 1  if result was zero, 0 otherwise
N = 1  if bit 7 of the result is 1, 0 otherwise
V = 1  if bit 7 of the accumulator was changed, a sign change``````

Increment and Decrement Instructions

INC – INCrement memory by one
INX – INcrement X by one
INY – INcrement Y by one

DEC – DECrement memory by one
DEX – DEcrement X by one
DEY – DEcrement Y by one

The 6502 has instructions for incrementing/decrementing the index
registers and memory. Note that it does not have instructions for
incrementing/decrementing the accumulator. This oversight was
rectified in the 65C02 which added INA and DEA instructions. The
index register instructions are implied mode for obvious reasons while
the INC and DEC instructions use a number of addressing modes.

All inc/dec instructions have alter the processor status flags in the
following way:

`````` Z = 1  if the result is zero, 0 otherwise
N = 1  if bit 7 is 1, 0 otherwise``````

Logical Instructions

AND – AND memory with accumulator
ORA – OR memory with Accumulator
EOR – Exclusive-OR memory with Accumulator

These instructions perform a bitwise binary operation according to the
tables given above. They set the Z flag if the net result is zero and
set the N flag if bit 7 of the result is set.

Jump, Branch, Compare, and Test Bits

BCC – Branch on Carry Clear, C = 0
BCS – Branch on Carry Set, C = 1
BEQ – Branch on EQual to zero, Z = 1
BNE – Branch on Not Equal to zero, Z = 0
BMI – Branch on MInus, N = 1
BPL – Branch on PLus, N = 0
BVS – Branch on oVerflow Set, V = 1
BVC – Branch on oVerflow Clear, V = 0

CMP – CoMPare memory and accumulator
CPX – ComPare memory and X
CPY – ComPare memory and Y

BIT – test BITs

This large group includes all instructions that alter the flow of the
program or perform a comparison of values or bits.

JMP simply sets the program counter (PC) to the address given.
Execution proceeds from the new address. The branch instructions are
relative jumps. They cause a branch to a new address that is either
127 bytes beyond the current PC or 128 bytes before the current PC.
Code that only uses branch instructions is relocatable and can be run
anywhere in memory.

The three compare instructions are used to set processor status bits.
After the comparison one frequently branches to a new place in the
program based on the settings of the status register. The
relationship between the compared values and the status bits is,

``````      +-------------------------+---------------------+
|                         |  N       Z       C  |
+-------------------------+---------------------+
| A, X, or Y  <  Memory   |  1       0       0  |
| A, X, or Y  =  Memory   |  0       1       1  |
| A, X, or Y  >  Memory   |  0       0       1  |
+-----------------------------------------------+``````

The BIT instruction tests bits in memory with the accumulator but
changes neither. Only processor status flags are set. The contents
of the specified memory location are logically ANDed with the
accumulator, then the status bits are set such that,

• N receives the initial, un-ANDed value of memory bit 7.
• V receives the initial, un-ANDed value of memory bit 6.
• Z is set if the result of the AND is zero, otherwise reset. So, if \$23 contained \$7F and the accumulator contained \$80 a BIT \$23
instruction would result in the V and Z flags being set and N reset since
bit 7 of \$7F is 0, bit 6 of \$7F is 1, and \$7F AND \$80 = 0. Shift and Rotate Instructions ASL – Accumulator Shift Left
LSR – Logical Shift Right
ROL – ROtate Left
ROR – ROtate Right Use these instructions to move things around in the accumulator or
memory. The net effects are (where C is the carry flag): `+-+-+-+-+-+-+-+-+` C <- |7|6|5|4|3|2|1|0| <- 0 ASL
+-+-+-+-+-+-+-+-+ `+-+-+-+-+-+-+-+-+` 0 -> |7|6|5|4|3|2|1|0| -> C LSR
+-+-+-+-+-+-+-+-+ `+-+-+-+-+-+-+-+-+` C <- |7|6|5|4|3|2|1|0| <- C ROL
+-+-+-+-+-+-+-+-+ `+-+-+-+-+-+-+-+-+` C -> |7|6|5|4|3|2|1|0| -> C ROR
+-+-+-+-+-+-+-+-+ Z is set if the result it zero. N is set if bit 7 is 1. It is
always reset on LSR. Remember that ASL A is equal to multiplying by
two and that LSR is equal to dividing by two. Transfer Instructions TAX – Transfer Accumulator to X
TAY – Transfer Accumulator to Y
TXA – Transfer X to accumulator
TYA – Transfer Y to Accumulator Transfer instructions move values between the 6502 registers. The N
and Z flags are set if the value being moved warrants it, i.e. LDA #\$80
TAX causes the N flag to be set since bit 7 of the value moved is 1, while LDX #\$00
TXA causes the Z flag to be set since the value is zero. Stack Instructions TSX – Transfer Stack pointer to X
TXS – Transfer X to Stack pointer PHA – PusH Accumulator on stack
PHP – PusH Processor status on stack
PLA – PulL Accumulator from stack
PLP – PulL Processor status from stack TSX and TXS make manipulating the stack possible. The push and pull
instructions are useful for saving register values and status flags.
RTS – ReTurn from Subroutine
RTI – ReTurn from Interrupt Like JMP, JSR causes the program to start execution of the next
of the next instruction after itself on the stack. When an RTS
instruction is executed the address pushed on the stack is pulled off
the stack and the program resumes at that address. For example, LDA #\$C1 ; load the character ‘A’
JSR print ; print the character and it’s hex code
JSR print ; and print it
.
.
.
print JSR \$FDED ; print the letter
JSR \$FDDA ; and its ASCII code
RTS ; return to the caller RTI is analagous to RTS and should be used to end an interrupt routine. Set and Reset (Clear) Instructions CLC – CLear Carry flag
CLD – CLear Decimal mode
CLI – CLear Interrupt disable
CLV – CLear oVerflow flag SEC – SEt Carry
SED – SEt Decimal mode
SEI – SEt Interrupt disable These are one byte instructions to specify processor status flag
settings. CLC and SEC are of particular use in addition and subtraction
or the result may be one greater than you expect. For subtraction
(SBC) use SEC to ensure that the carry is set as its compliment is
only clear or set the carry flag before the initial operation. For
example, to add one to a 16-bit number in \$23 and \$24 you would write: LDA \$23 ; get the low byte
CLC ; clear the carry
ADC #\$02 ; add a constant 2, carry will be set if result > 255
STA \$23 ; save the low byte
LDA \$24 ; get the high byte
STA \$24 ; save the high byte
RTS ; if carry set now the result was > 65535 Similarly for subtraction, LDA \$23 ; get the low byte
SEC ; set the carry
SBC #\$02 ; subtract 2
STA \$23 ; save the low byte
LDA \$24 ; get the high byte
SBC #\$00 ; subtract 0 and any borrow generated above
STA \$24 ; save the high byte
RTS ; if the carry is not set the result was < 0 Other Instructions NOP – No OPeration (or is it NO oPeration ? 🙂
BRK – BReaK NOP is just that, no operation. Useful for deleting old
instructions, reserving room for future instructions or for use in
careful timing loops as it uses 2 microprocessor cycles. BRK causes a forced break to occur and the processor will immediately
start execution of the routine whose address is in \$FFFE and \$FFFF.
This address is often the start of a system monitor program.

# Some simple programming examples

``````A few simple programming examples are given here.  They serve to
illustrate some techniques commonly used in assembly programming.
There are doubtless dozens more and I make no claim at being a
proficient assembly language programmer.  For examples of addition
and subtraction see above on CLC and SEC.

A count down loop
-----------------

;
; An 8-bit count down loop
;

start LDX #\$FF    ; load X with \$FF = 255
loop  DEX         ; X = X - 1
BNE loop    ; if X not zero then goto loop
RTS         ; return

How does the BNE instruction know that X is zero?  It
doesn't, all it knows is that the Z flag is set or reset.
The DEX instruction will set the Z flag when X is zero.

;
; A 16-bit count down loop
;

start LDY #\$FF    ; load Y with \$FF
loop1 LDX #\$FF    ; load X with \$FF
loop2 DEX         ; X = X - 1
BNE loop2   ; if X not zero goto loop2
DEY         ; Y = Y - 1
BNE loop1   ; if Y not zero goto loop1
RTS         ; return

There are two loops here, X will be set to 255 and count to
zero for each time Y is decremented.  The net result is to
count the 16-bit number Y (high) and X (low) down from \$FFFF
= 65535 to zero.

Other examples
--------------

** Note: All of the following examples are lifted nearly verbatim from
the book "6502 Software Design", whose reference is above.

; Example 4-2.  Deleting an entry from an unordered list
;
; Delete the contents of \$2F from a list whose starting
; address is in \$30 and \$31.  The first byte of the list
; is its length.
;

deluel  LDY #\$00     ; fetch element count
LDA (\$30),Y
TAX          ; transfer length to X
LDA \$2F      ; item to delete
nextel  INY          ; index to next element
CMP (\$30),Y  ; do entry and element match?
BEQ delete   ; yes. delete element
DEX          ; no. decrement element count
BNE nextel   ; any more elements to compare?
RTS          ; no. element not in list. done

; delete an element by moving the ones below it up one location

delete  DEX          ; decrement element count
BEQ deccnt   ; end of list?
INY          ; no. move next element up
LDA (\$30),Y
DEY
STA (\$30),Y
INY
JMP delete
deccnt  LDA (\$30,X)  ; update element count of list
SBC #\$01
STA (\$30,X)
RTS

; Example 5-6.  16-bit by 16-bit unsigned multiply
;
; Multiply \$22 (low) and \$23 (high) by \$20 (low) and
; \$21 (high) producing a 32-bit result in \$24 (low) to \$27 (high)
;

mlt16   LDA #\$00     ; clear p2 and p3 of product
STA \$26
STA \$27
LDX #\$16     ; multiplier bit count = 16
nxtbt   LSR \$21      ; shift two-byte multiplier right
ROR \$20
BCC align    ; multiplier = 1?
LDA \$26      ; yes. fetch p2
CLC
STA \$26      ; store new p2
LDA \$27      ; fetch p3
align   ROR A        ; rotate four-byte product right
STA \$27      ; store new p3
ROR \$26
ROR \$25
ROR \$24
DEX          ; decrement bit count
BNE nxtbt    ; loop until 16 bits are done
RTS

; Example 5-14.  Simple 16-bit square root.
;
; Returns the 8-bit square root in \$20 of the
; 16-bit number in \$20 (low) and \$21 (high). The
; remainder is in location \$21.

sqrt16  LDY #\$01     ; lsby of first odd number = 1
STY \$22
DEY
STY \$23      ; msby of first odd number (sqrt = 0)
again   SEC
LDA \$20      ; save remainder in X register
TAX          ; subtract odd lo from integer lo
SBC \$22
STA \$20
LDA \$21      ; subtract odd hi from integer hi
SBC \$23
STA \$21      ; is subtract result negative?
BCC nomore   ; no. increment square root
INY
LDA \$22      ; calculate next odd number
STA \$22
BCC again
INC \$23
JMP again
nomore STY \$20      ; all done, store square root
STX \$21      ; and remainder
RTS

This is based on the observation that the square root of an
integer is equal to the number of times an increasing odd
number can be subtracted from the original number and remain
positive.  For example,

25
-  1         1
--
24
-  3         2
--
21
-  5         3
--
16
-  7         4
--
9
-  9         5 = square root of 25
--
0``````
```

```