Example Programs

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Example Programs

Programming is at once a skill and an art. Just as anyone may learn to play a musical instrument after sufficient instruction and practice. so may anyone learn to program a computer. Some individuals. however. have a gift for programming that sets them apart from their peers with the same level of experience. just as some musicians are more tal­ented than their contemporaries.

Gifted or not. you will not become adept at programming until you have written and rewritten many programs. The emphasis here is on practice; you can read many books on how to ride a bicycle, but you do not know how to ride until you do it.

If some of the examples and problems seem trivial or without any "real-world" appli­cation. remember the playing of scales on a piano by a budding musician. Each example will be done using several methods; the best method depends upon what resource is in short supply. If programming time is valuable. then the best program is the one that uses the fewest lines of code; if either ROM or execution time is limited. then the program that uses the fewest code bytes is best.

¨ EXAMPLE PROBLEM 3.1

Copy the byte in TCON to register R2 using at least four different methods.

· Method1: Use the direct address for TCON (88h) and register R2.

Mnemonic Operation

MOV R2.88h Copy TCON to R2

· Method 2: Use the direct addresses for TCON and R2.

Mnemonic Operation

MOV 02h.88h Copy TCON to direct address 02h (R2)

· Method 3: Use R1 as a pointer to R2 and use the address of TCON.

Mnemonic Operation

MOV R1,#02h Use R1 as a pointer to R2

MOV @R1 .88h Copy TCON byte to address in R1 (02h = R2)

· Method 4: Push the contents of TCON into direct address 02h (R2).

Mnemonic Operation

MOV 81h.#01h Set the SP to address 01 h in RAM

PUSH 88h Push TCON (88h) to address 02h (R2)

¨ EXAMPLE PROBLEM 3.2

Set timer T0 to an initial setting of 1234h.

· Method 1: Use the direct address with an immediate number to set THO and TLO.

Mnemonic Operation

MOV 8Ch,#12h Set TH0 to 12h

MOV 8Ah,#34h Set TL0 to 34h

Totals: 6 bytes, 2 lines

· Method 2: Use indirect addressing with RO for TLO and RI for THO.

Mnemonic Operation

MOV R0,#8Ah Copy 8Ah, the direct address of TL0, to R0

MOV R1,#8Ch Copy 8Ch, the direct address of TH0, to R1

MOV @R0,#34h Copy 34h to TLO

MOV @R1,#12h Copy 12h to THO

Totals: 8 bytes, 4 lines

The first method is also the better method in this example.

¨ EXAMPLE PROBLEM 3.3

Put the number 34h in registers R5, R6, and R7.

· Method 1: Use an immediate number and register addressing.

Mnemonic Operation

MOV R5,#34h Copy 34h to R5

MOV R6,#34h Copy 34h to R6

MOV R7,#34h Copy 34h to R7

Totals: 6 bytes, 3 lines

· Method 2: Since the number is the same for each register, put the number in A and MOV A to each register.

Mnemonic Operation

MOV A,#34h Copy a 34h to A

MOV R5,A Copy A to R5

MOV R6,A Copy A to R6

MOV R7,A Copy A to R7

Totals: 5 bytes, 4 lines

· Method 3: Copy one direct address to another.

Mnemonic Operation

MOV R5,#34h Copy 34h to register R5

MOV 06h,05h Copy R5 (add 05) to R6 (add 06)

MOV 07h,06h Copy R6 to R7

Totals: 8 bytes, 3 lines

¨ EXAMPLE PROBLEM 3.4

Put the number 80h in RAM locations 30h to 34h.

· Method 1: Use the immediate number to a direct address:

Mnemonic Operation

MOV 30h,#8Dh Copy the number 80h to RAM address 30h

MOV 31h,#8Dh Copy the number 80h to RAM address 31 h

MOV 32h,#8Dh Copy the number 8Dh to RAM address 32h

MOV 33h,#8Dh Copy the number 80h to RAM address 33h

MOV 34h,#8Dh Copy the number 80h to RAM address 34h

Totals: 15 bytes,5 lines

· Method 2: Using the immediate number in each instruction uses bytes; use a register to hold the number:

Mnemonic Operation

MOV A,#80h Copy the number 80h to the A register

MOV 30h,A Copy the contents of A to RAM address 30h

MOV 31h,A Copy the contents of A to the remaining addresses

MOV 32h,A

MOV 33h,A

MOV 34h,A Totals: 12 bytes, 6 lines

· Method 3: There must be a way to avoid naming each address; the PUSH opcode can increment to each address:

Mnemonic Operation

MOV 30h,#8Dh Copy the number 8Dh to RAM address 30h

MOV 81h,#3Dh Set the SP to 30h

PUSH 30h Push the contents of 30h (=8Dh) to address 31h

PUSH 30h Continue pushing to address 34h

PUSH 30h

PUSH 30h Totals: 14 bytes, 6 lines

COMMENT

Indirect addressing with the number in A and the indirect address in R 1 could be done; how­ever, R 1 would have to be loaded with each address from 30h to 34h. Loading R 1 would take a total of 17 bytes and 11 lines of code. Indirect addressing is advantageous when we have opcodes that can change the contents of the pointing registers automatically.

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