Read two numbers and print their sum

This program asks for two numbers and prints their sum. Each prompt uses task 18 of trap #15: the task prints the zero-terminated string at a1, waits for you to enter a number, and returns that number in d1.l.

    lea first, a1
    move.b #18, d0      ; print the first prompt and read a number
    trap #15
    move.l d1, d2       ; save the first number in d2

    lea second, a1
    move.b #18, d0      ; print the second prompt and read a number
    trap #15
    add.l d1, d2        ; add the second number to the saved first number

    lea answer, a1
    move.l d2, d1       ; task 17 prints the number in d1
    move.b #17, d0      ; print the answer string and the sum
    trap #15

    move.b #9, d0       ; end the program
    trap #15

    org $2000
first:  dc.b 'First number: ', 0
second: dc.b 10, 'Second number: ', 0
answer: dc.b 10, 'The sum is ', 01
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22

Select Build, then Run. At First number: , type 17 in the console, where the caret waits, and press Enter. The program continues to Second number: and waits again. Type 25 and press Enter; it prints The sum is 42.

The first task 18 returns 17 in d1.l. move.l d1, d2 keeps a copy because the next task 18 puts its answer, 25, in d1.l. Then add.l d1, d2 adds that second answer directly to the saved first number: d2.l becomes 42. Task 17 needs its number in d1.l, so move.l d2, d1 puts the sum there for printing.

The 10 before Second number: and The sum is is the ASCII code for a newline. Each string starts on a new line when its task prints it. The final 0 marks the end of each string.

Use small whole numbers for this example. The arithmetic uses 32-bit registers, so a sum that does not fit in 32 bits will wrap around instead of showing the mathematical result.

Predict the result

Build and run again with 8 as the first input and -3 as the second. What will d2.l hold after add.l, and what will the program print?

Show answer

d2.l holds 5, and the program prints The sum is 5. The first input is saved in d2.l; adding the second input from d1.l gives 8 + (-3) = 5.