http://courses.cs.vt.edu/csonline/MachineArchitecture/Lessons/CPU/
http://www.emu8086.com/assembler_tutorial/tutorials.html

C code: C程式碼
	n = 10;
	n = n+23;

6502 assembler: 組合語言(8 bit)
	0500	lda #23		Load載入 the byte value 23h into the accumulator累加器(ACC=23)
	0502	add $2043	Add加 the byte of data found at memory 2043h (ACC=ACC+n)
	0505	sta $2043	Store儲存於 at memory 2043h (n=ACC)
	..
	2043	10		(int n)

As 6502 machine code in memory: 機器碼
	9a 23 8a 43 20 84 44 20 ..

Note that 6502 uses little-endian, so adc $2043 is  8a 43 20


lda #23 (2 cycles)
Fetch:	PC=0500		// addr of next instruction
	MAR=PC		// memory address register
	MDR=memory[MAR]	// memory data register
	IR=MDR		// instruction(lda)
	PC++

Decode:	operand (#23) needed
	MAR=PC		// PC=0501
	MDR=memory[MAR]	// operand #23
	PC++

Execute:
	ACC=MDR		// #23
	PSR updated


adc $2043 (4 cycles)
Fetch:	PC=0502
	MAR=PC
	MDR=memory[MAR]
	IR=MDR		// instruction(adc)
	PC++

Decode:	next 2 bytes are addresses (T=$2043)
	MAR=PC
	MDR=memory[MAR]	// addr 43
	T[1]=MDR
	PC++

	MAR=PC
	MDR=memory[MAR]	// addr 20
	T[0]=MDR
	PC++

Execute:
	MAR=T		// addr 2043
	MDR=memory[MAR]	// data(10)
	ACC=ACC+MDR	// 23+10
	PSR updated


sta $2043 (4 cycles)
Fetch:	PC=0505
	MAR=PC
	MDR=memory[MAR]
	IR=MDR	// instruction(sta)
	PC++

Decode:	next 2 bytes are addresses (T=$2043)
	MAR=PC
	MDR=memory[MAR]	// addr 43
	T[1]=MDR
	PC++

	MAR=PC
	MDR=memory[MAR]	// addr 20
	T[0]=MDR
	PC++

Execute:
	MAR=T		// addr 2043
	MDR=ACC		// memory[MAR]=MDR



Example 2:

for n in range(6): …
for(n=0;n<=5;n++)…

0001 LOAD #5	// ACC=5
0002 STORE 0020	// MAX=5
0003 LOAD #0	// ACC=0
0004 STORE 0030	// n=ACC
0005 SUB 0020	// ACC=ACC-5
0006 JPZ 0010	// if(ACC==0) jump to 0010 (STOP)
0007 LOAD 0030	// ACC=n
0008 ADD #1	// ACC++
0009 JMP 4
0010 STOP
…
0020 ?	// MAX
…
0030 ?	// n
…