Welcome to Data Crystal's new home! Data Crystal is now part of the TCRF family (sort of).
The wiki has recently moved; please report any issues in Discord. Pardon the dust.

User:HatZen08/draft: Difference between revisions

From Data Crystal
Jump to navigation Jump to search
(Personal Draft)
 
(To delete personal draft.)
 
Line 1: Line 1:
== Introduction ==
The article was deleted.
 
This is a list of the opcodes used in the 65816 processor. It is used by the Super Nintendo System (SNES) console.
 
For the registers, the following convention will be used:
 
Convention Description
A          the low byte of the accumulator, always 8-bit
B          the high byte of the accumulator, always 8-bit
C          the low and high bytes of the accumulator, always 16-bit
X          the X register, 8-bit or 16-bit
Y          the Y register, 8-bit or 16-bit
P          the processor register, always 8-bit
 
For the opcode arguments, the following convention will be used:
 
Argument  Example    Relative to
addr      LDA $1234  memory address, short, 16-bit
const      LDA #$80    constant, 8-bit or 16-bit
dp        LDA $10    direct page, relative, 8-bit
long      LDA $C21234 memory address, long, 24-bit
near      BRA $1234  branch, relative, 8-bit
sr        LDA $01,S  stack, relative, 8-bit
 
== LDA, LDX, LDY ==
 
Reads a value from the memory and stores it in the specified register. LDA will store in the A or C register, LDX will store in the X register and LDY will store in the Y register.
 
flags
n highest bit of value loaded.
z set if data loaded is zero.
 
Opcode Syntax          Bytes  Notes
A9    LDA const      2      One extra byte if in 16-bit mode
AD    LDA addr        3
AF    LDA long        4
A5    LDA dp          2
B2    LDA (dp)        2
A7    LDA [dp]        2
BD    LDA addr,X      3
BF    LDA long,X      4
B9    LDA addr,Y      3
B5    LDA dp,X        2
A1    LDA (dp,X)      2
B1    LDA (dp),Y      2
B7    LDA [dp],Y      2
A3    LDA sr,S        2
B3    LDA (sr,S),Y    2
 
Opcode Syntax          Bytes  Notes
A2    LDX const      2      One extra byte if in 16-bit mode
AE    LDX addr        3
A6    LDX dp          2
BE    LDX addr,Y      3
B6    LDX dp,Y        2
 
Opcode Syntax          Bytes  Notes
A0    LDY const      2      One extra byte if in 16-bit mode
AC    LDY addr        3
A4    LDY dp          2
BC    LDY addr,X      3
B4    LDY dp,X        2
 
== STA, STX, STY ==
 
Reads a value from the specified register and stores it in the memory. STA will read the A or C register, STX will read the X register and STY will read the Y register. All opcodes will store the retrieved value in the specified address.
 
The STZ opcode will always store the value of zero in the designed memory address. Unlike the STA, STX and STY opcodes, the STZ opcode doesn't affect flags. It also doesn't affect the A or C register.
 
flags
??
Opcode Syntax          Bytes  Notes
8D    STA addr        3
8F    STA long        4
85    STA dp          2
92    STA (dp)        2
87    STA [dp]        2
9D    STA addr,X      3
9F    STA long,X      4
99    STA addr,Y      3
95    STA dp,X        2
81    STA (dp,X)      2
91    STA (dp),Y      2
97    STA [dp],Y      2
83    STA sr,S        2
93    STA (sr,S),Y    2
 
Opcode Syntax          Bytes  Notes
8E    STX addr        3
86    STX dp          2
96    STX dp,Y        2
 
Opcode Syntax          Bytes  Notes
8C    STY addr        3
84    STY dp          2
94    STY dp,X        2 
 
Opcode Syntax          Bytes  Notes
9C    STZ addr        3
64    STZ dp          2
9E    STZ addr,X      3
74    STZ dp,X        2
 
== TRB, TSB ==
 
Test all bits in the A or C register. For all bits set in the A or C register, the correspondent bits will be set or unset in the designed memory address. TSB will set the correspondent bits and TRB will reset the correspondent bits. All clear bits in the A or C register will be ignored and their correspondent bits in the designed address will not be altered.
 
flags
??
 
Opcode Syntax          Bytes  Notes
1C    TRB addr        3
14    TRB dp          2
 
Opcode Syntax          Bytes  Notes
0C    TSB addr        3
04    TSB dp          2
 
== ADC, SBC ==
 
Add or subtract the value in the A or C register with the value in the designed memory address. ADC will add the value and SBC will subtract the value.
 
When the carry flag is set, a value of one will be added in the addition or subtraction operation.
 
The operation can overflow or underflow. It is the coder responsibility to check these cases and adjust the resulting value. A common method is to check the carry flag for overflow or underflow and cap the value to a prefixed setting.
 
flags:
n set if highest bit of result was set
v set if signed overflow?
z set if result is zero
c set if overflow or underflow
 
Opcode Syntax          Bytes  Notes
61    ADC (dp,X)      2
63    ADC sr,S        2
65    ADC dp          2
67    ADC [dp]        2
69    ADC const      2      One extra byte if in 16-bit mode
6D    ADC addr        3     
6F    ADC long        4
71    ADC (dp),Y      2
72    ADC (dp)        2
73    ADC (sr,S),Y    2
75    ADC dp,X        2
77    ADC [dp],Y      2
79    ADC addr,Y      3
7D    ADC addr,X      3
7F    ADC long,X      4
 
Opcode Syntax          Bytes  Notes
E9    SBC const      2      One extra byte if in 16-bit mode
ED    SBC addr        3
EF    SBC long        4
E5    SBC dp          2
F2    SBC (dp)        2
E7    SBC [dp]        2
FD    SBC addr,X      3
FF    SBC long,X      4
F9    SBC addr,Y      3
F5    SBC dp,X        2
E1    SBC (dp,X)      2
F1    SBC (dp),Y      2
F7    SBC [dp],Y      2
E3    SBC sr,S        2
F3    SBC (sr,S),Y    2
 
== INC, INX, INY, DEC, DEX, DEY ==
 
Increment or decrement the specified register by one. It is equivalent to the ADC or SBC opcodes with the value of one without the carry flag set.
 
INC will increment the A or C register. INX will increment the X register. INY will increment the Y register. DEC will decrement the A or C register. DEX will decrement the X register. DEY will decrement the Y register. The operation can underflow or overflow.
 
flags:
n set if most significant bit of result is set.
z set if result is zero.
 
Opcode Syntax          Bytes  Notes
1A    INC            1      increment the A register
EE    INC addr        3
E6    INC dp          2
FE    INC addr,X      3
F6    INC dp,X        2
E8    INX            1      increment the X register
C8    INY            1      increment the Y register
 
Opcode Syntax          Bytes  Notes
3A    DEC            1      decrement register A
CE    DEC addr        3
C6    DEC dp          2
DE    DEC addr,X      3
D6    DEC dp,X        2
CA    DEX            1      decrement register X
88    DEY            1      decrement register Y
 
== AND, ORA, EOR ==
 
A binary operation is done between the register A or C and the memory. The AND opcode will do a binary AND, the ORA opcode will do a binary OR and the EOR opcode will do a binary XOR. The result will be stored in the A or C register.
 
Flags:
n set if highest bit is set
z set if result is zero
 
Opcode Syntax          Bytes  Notes
21    AND (dp,X)      2
23    AND sr,S        2
25    AND dp          2
27    AND [dp]        2
29    AND const      2      One extra bit if in 16-bit mode
2D    AND addr        3
2F    AND long        4
31    AND (dp),Y      2
32    AND (dp)        2
33    AND (sr,S),Y    2
35    AND dp,X        2
37    AND [dp],Y      2
39    AND addr,Y      3
3D    AND addr,X      3
3F    AND long,X      4
 
Opcode Syntax          Bytes  Notes
09    ORA const      2      One extra byte if in 16-bit mode
0D    ORA addr        3
0F    ORA long        4
05    ORA dp          2
12    ORA (dp)        2
07    ORA [dp]        2
1D    ORA addr,X      3
1F    ORA long,X      4
19    ORA addr,Y      3
15    ORA dp,X        2
01    ORA (dp,X)      2
11    ORA (dp),Y      2
17    ORA [dp],Y      2
03    ORA sr,S        2
13    ORA (sr,S),Y    2
 
Opcode Syntax          Bytes  Notes
49    EOR const      2      One extra byte if in 16-bit mode
4D    EOR addr        3
4F    EOR long        4
45    EOR dp          2
52    EOR (dp)        2
47    EOR [dp]        2
5D    EOR addr,X      3
5F    EOR long,X      4
59    EOR addr,Y      3
55    EOR dp,X        2
41    EOR (dp,X)      2
51    EOR (dp),Y      2
57    EOR [dp],Y      2
43    EOR sr,S        2
53    EOR (sr,S),Y    2
 
== BIT ==
 
Does a binary AND between the A or C register and the memory. Unlike the AND opcode, the binary AND operation only is done for the bits sets in the A or C register. All clear bits in the A or C register will be left clear after the operation.
 
flags:
n Takes value of most significant bit of memory data.
v Takes value from bit 6 or 14 of memory data.
z Set if logical AND of mem and acc is zero.
 
Opcode Syntax          Bytes  Notes
89    BIT const      2      Add one extra bit if in 16-bit mode
2C    BIT addr        3
24    BIT dp          2
3C    BIT addr,X      3
34    BIT dp,X        2
 
== CMP, CPX, CPY ==
 
Compare the designed register with the memory and sets flags based on the comparison. CMP will compare the A or C register, CPX will compare the X register and CPY will compare the Y register. Generally, a conditional branch opcode, based on a flag status, is used after these opcodes.
 
flags:
n set if most significant bit of result is set
z set if result is zero
c set if register is equal or greater than memory
Opcode Syntax          Bytes  Notes
C9    CMP const      2      One extra byte if in 16-bit mode
CD    CMP addr        3
CF    CMP long        4
C5    CMP dp          2
D2    CMP (dp)        2
C7    CMP [dp]        2
DD    CMP addr,X      3
DF    CMP long,X      2
D9    CMP addr,Y      3
D5    CMP dp,X        2
C1    CMP (dp,X)      2
D1    CMP (dp),Y      2
D7    CMP [dp],Y      2
C3    CMP sr,S        2
D3    CMP (sr,S),Y    2
 
Opcode Syntax          Bytes  Notes
E0    CPX const      2      One extra byte if in 16-bit mode
EC    CPX addr        3
E4    CPX dp          2
Opcode Syntax          Bytes  Notes
C0    CPY const      2      One extra byte if in 16-bit mode
CC    CPY addr        3
C4    CPY dp          2
 
== LSR, ASL ==
 
Shift bits in the designed register or memory address left or right. The LSR opcode will shift the bits right and the ASL opcode will shift the bits left.
 
LSR is equivalent to divide the number by two and ASL is equivalent to multiply the number by two. The operation can underflow or overflow. The bit shifted out will become the new carry. The bit shifted in is zero.
 
flags: LSR
n cleared.
z set if result is zero.
c bit zero becomes new carry.
 
flags: ASL
n set if most significant bit of result is set.
z set if result is zero.
c set if highest bit is moved into carry
 
Opcode Syntax          Bytes  Notes
4A    LSR            1      shift A register
4E    LSR addr        3
46    LSR dp          2
5E    LSR addr,X      3
56    LSR dp,X        2
 
Opcode Syntax          Bytes  Notes
06    ASL dp          2
0A    ASL            1      The operand is the register A
0E    ASL addr        3
16    ASL dp,X        2
1E    ASL addr,X      3
 
== ROL, ROR ==
 
Shift bits in the designed register or memory address left or right. The ROL opcode will shift the bits left and the ROR opcode will shift the bits right.
 
The bit shifted in is the old carry. The bit shifted out will be the new carry.
 
flags: ROL
n set if most significant bit of result is set.
z set if result is zero.
c the high bit (7 or 15) becomes the new carry.
 
flags: ROR
n set if most significant bit of result is set.
z set if result is zero.
c low bit becomes the new carry.
 
Opcode Syntax          Bytes  Notes
2A    ROL            1      affects register A
2E    ROL addr        3
26    ROL dp          2
3E    ROL addr,X      3
36    ROL dp,X        2
 
Opcode Syntax          Bytes  Notes
6A    ROR            1      affects A register
6E    ROR addr        3
66    ROR dp          2
7E    ROR addr,X      3
76    ROR dp,X        2
 
== BPL, BMI, BRA, BCC, BCS, BNE, BEQ, BVC, BVS, BRL ==
 
Change the execution flow based on a conditional branch. If the condition is meet, the execution flow will be changed for the designed address. Otherwise, it continues normally.
 
The majority of the conditional branches are based in flag status. The flags were setup by anterior opcodes.
 
flags:
none
 
Opcode Syntax          Bytes  Notes
10    BPL near        2      BPL - branch if minus flag is clear, branch if plus
30    BMI near        2      BMI - branch if minus flag set, branch if minus
80    BRA near        2      BRA - always branch
90    BCC near        2      BCC - branch if carry flag clear
B0    BCS near        2      BCS - branch if carry flag set
D0    BNE near        2      BNE - branch if zero flag clear, branch if not equal
F0    BEQ near        2      BEQ - branch if zero flag set, branch if equal
50    BVC near        2      BVC - branch if overflow clear
70    BVS near        2      BVS - branch if overflow set
82    BRL addr        3      similar to BRA, but with longer range
 
== JMP ==
 
Change the execution flow to the designed address.
 
flags:
none
 
Opcode Syntax          Bytes  Notes
4C    JMP addr        3
6C    JMP (addr)      3
7C    JMP (addr,X)    3
5C    JMP long        4
DC    JMP [addr]      3
 
== JSR, JSL ==
 
Calls a sub routine which will be executed.
 
The JSR opcode will put two bytes in the stack, relative to the return address. It is expected that they will be retrieved by an RTS opcode in the sub routine.
 
The JSL opcode will put three bytes in the stack, relative to the return address. It is expected that they will be retrieved by an RTL opcode in the sub routine.
 
Opcode Syntax          Bytes  Notes
22    JSL long        4
20    JSR addr        3
FC    JSR (addr,X)    3
 
== RTS, RTL ==
 
RTS will pull two bytes from the stack and RTL will pull three bytes from the stack. They will be setup as the new control flow address.
 
It is expected that the bytes in the stack were pushed from a JSR opcode for the RTS opcode and from a JSL opcode for the RTL opcode.
 
flags:
none
 
Opcode Syntax          Bytes  Notes
6B    RTL            1      pull three bytes from the stack as the return address
60    RTS            1      pull two bytes from the stack as the return address
 
== CLC, CLD, CLI, CLV, SEC, SED, SEI ==
 
Set or reset flags.
 
Opcode Syntax          Bytes  Notes
18    CLC            1      clear carry flag
D8    CLD            1      clear decimal flag
58    CLI            1      clear interrupt flag
B8    CLV            1      clear overflow flag
38    SEC            1      set carry flag
F8    SED            1      set decimal flag
78    SEI            1      set interrupt flag
 
== TAX, TAY, TXA, TYA, TSX, TXS, TXY, TYX, TCD, TDC, TCS, TSC ==
 
Transfer bytes between the specified registers or special addresses.
 
If the setting of the registers have different settings of 8-bit or 16-bit, the copy of B can be truncated based on the opcode behavior. It is possible that a leftover value of B will be left untouched in the B register.
 
flags: all except TCS
n set if most significant bit of transfer value is set.
z set if transferred value is zero.
 
Opcode Syntax          Bytes  Notes
AA    TAX            1      from A or C to X register
A8    TAY            1      from A or C to Y register
8A    TXA            1      from X register to A or C
98    TYA            1      from Y register to A or C
BA    TSX            1      from stack pointer to X register
9A    TXS            1      from X register to stack pointer
9B    TXY            1      from X register to Y register
BB    TYX            1      from Y register to X register
5B    TCD            1      from C to direct page
7B    TDC            1      from direct page to C
1B    TCS            1      from C to stack pointer
3B    TSC            1      from stack pointer to C
 
== MVP, MVN ==
 
Move a block of bytes from a specified address to another address. The X register will be the start address to copy from. The Y register will be start address to copy to. The C register will be the number of bytes to copy minus one.
 
The opcode arguments are the source bank and the destination bank, in this order.
 
MVP must be used if the Y register is greater than the X register. Otherwise, MVN must be used instead.
 
Opcode Syntax          Bytes  Notes
54    MVN src,dest    3
44    MVP src,dest    3
 
== PEA, PEI, PER ==
 
Push bytes in the stack, based on the opcodes arguments. They always work in 16-bit mode.
 
Opcode Syntax          Bytes  Notes
F4    PEA const      3      the argument is always 16-bit
D4    PEI (dp)        2
62    PER label      3
 
== PHA, PHP, PHX, PHY, PHB, PHD, PHK, PLA, PLP, PLX, PLY, PLB, PLD ==
 
Push or pull bytes from the stack. The setting of the register or special address as 8-bit or 16-bit will set the number of bytes pushed/pulled from the stack.
 
flags: PHA/PHP/PHX/PHY/PHB/PHD/PHK
none
 
flags: PLA/PLX/PLY/PLB/PLD
n set if most significant bit of result is set.
z set if result is zero.
 
Opcode Syntax          Bytes  Notes
48    PHA            1      push A or C register
08    PHP            1      push flag register (8-bit)
DA    PHX            1      push X register (8-bit or 16-bit)
5A    PHY            1      push Y register (8-bit or 16-bit)
8B    PHB            1      push data bank register (8-bit)
0B    PHD            1      push direct page register (16-bit)
4B    PHK            1      push program bank register (8-bit)
 
Opcode Syntax          Bytes  Notes
68    PLA            1      pull A or C register
28    PLP            1      pull flag register (8-bit)(sets all flags)
FA    PLX            1      pull X register (8-bit or 16-bit)
7A    PLY            1      pull Y register (8-bit or 16-bit)
AB    PLB            1      pull pull data bank register (8-bit)
2B    PLD            1      pull direct page register (16-bit)           
 
== NOP ==
 
Does absolutely nothing. It is mostly used to waste cycles. A few special registers or addresses can't be fetch until a minimum number of cycles has passed.
 
If the coder needs to erase a code segment, a conditional branch or jump is recommended instead of a long line of NOP opcodes.
 
Opcode Syntax          Bytes  Notes
EA    NOP            1
 
== REP, SEP ==
 
Setups the processor flags with the opcode arguments. For each bit set in the argument, the correspondent flag is set or reset. REP will reset the flags and SEP will set the flags.
 
arguments:
#$01 carry flag
#$02 zero flag
#$04 irq flag?
#$08 decimal mode
#$10 X and Y register: 0 = 16-bit, 1 = 8-bit
#$20 accumulator: 0 = 16-bit, 1 = 8-bit
#$40 overflow flag
#$80 negative flag
 
flags:
all
 
Opcode Syntax          Bytes  Notes
C2    REP const      2      const is always 8-bit
E2    SEP const      2      const is always 8-bit
 
== XBA ==
 
Exchanges the A and B registers.
 
flags:
n Set if the most significant bit of the new value in the low order 8 bits (A) of the accumulator is set. (former bit 15)
z Set if new value of the lower order 8 bit accumulator (A) is zero.
 
Opcode Syntax          Bytes  Notes
EB    XBA            1
 
== XCE ==
 
Exchanges the carry flag and the emulation flag.
 
flags:
e from previous carry flag.
c from previous emulation flag.
m native mode flag only. switching to native 65816 mode sets to one.
x x is a native mode flag only.
b brk is an emulation 6502 flag only. it is set to 1 to become the x flag in native mode
 
Opcode Syntax          Bytes  Notes
FB    XCE            1
 
== Miscellaneous ==
 
Opcode Syntax          Bytes  Notes
00    BRK            ??      software break
02    COP const      ??      coprocessor empowerment
40    RTI            1      return from interrupt
DB    STP            1      stop processor
CB    WAI            1      wait for interrupt

Latest revision as of 16:49, 21 December 2015

The article was deleted.

Retrieved from "https://datacrystal.tcrf.net/w/index.php?title=User:HatZen08/draft&oldid=34937"