Variables in memory and constants

This program adds a price of 250, shipping of 35, and tax of 20. The total is 305. It reads the price and shipping from memory, holds the running total in d0, then writes the answer back to memory.

TAX equ 20

    move.l price, d0        ; start with the price
    add.l shipping, d0      ; add shipping
    add.l #TAX, d0          ; add the tax value
    move.l d0, total        ; store the answer

    org $2000
price:    dc.l 250
shipping: dc.l 35
total:    ds.l 11
2
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11

price, shipping, and total name memory addresses. Although the instructions use those names before their declarations appear, Build reads the whole program and finds each address. The first instruction reads the long at price into d0; the last writes d0 to the long at total.

TAX equ 20 gives the number 20 a name. In add.l #TAX, d0, the # means the instruction uses that number directly. d0 holds 250, then 285, then 305 as the three instructions calculate the total.

The org $2000 line places the data that follows it at address $2000. The instructions above it stay at their earlier addresses. Each dc.l places one four-byte long in memory during Build, so shipping starts four bytes after price. ds.l 1 reserves one long, or four bytes, for total. It gives that space no initial value; the final move.l supplies the value when the program runs.

Run it and inspect memory

Select Build, then Run. In the memory panel, enter 2000 in the address box. Read four bytes at each address as one long:

label	address	value after Run
price	$2000	000000FA (250)
shipping	$2004	00000023 (35)
total	$2008	00000131 (305)

The result is hexadecimal $131, which is decimal 305. price and shipping keep the values placed there during Build; total receives the result during Run.

Try it

Change shipping from 35 to 40. Before selecting Build and Run again, predict the new value at total. Check it in the memory panel at $2008.

Show answer

The new total is 250 + 40 + 20 = 310, displayed as 00000136 in hexadecimal.