Read two numbers and print their sum

The program asks for two numbers, waits while you type them, and prints their sum. Press Run and the console stops at the first prompt: type a number into the box under it, press Enter, and the run carries on inside that one in instruction.

Print a string only talked. This one listens, which means the program stops in the middle of an instruction until somebody answers it, and what comes back is a number in a register rather than text you have to make sense of.

You need to know: the "Print a string" Example and the "Ports: in and out" lecture. What is new here is a port that is read, in a, (1) asks for a whole line and parses it as a decimal number, so the register the program reads next is the one the environment wrote.

Type 17 and 25 and the console reads The sum is 42. A line that is not a number stops the program with an error, and a number over 255 comes back as its low byte, because a is one byte and that is all the port can hand back.

Two bytes add up to as much as 510, which no byte holds, so the sum is worked out in sixteen bits: each number is widened with an ld h, 0 or an ld d, 0 as it arrives, and add hl, de adds the two pairs. Printing it takes the last port of the console set. Port 0x04 prints a 16 bit number whose low byte is what the out writes and whose high byte is the high byte of the address bus, and in the out (c), r form the address bus carries b, so ld b, h and ld c, 4 and out (c), l print the whole of hl.

The push hl and pop hl around the third call print are there because print walks hl to the end of the string it is printing. The sum has nowhere else to sit while that happens, and the stack is where a value waits for a call to finish.

The 10 at the front of second and answer is a newline written as its character code, which is how a line break goes inside a string. .db 10, "Second number: ", 0 is one string of seventeen bytes and the first of them is that break.

Try changing ld c, P_WORD to ld c, P_NUM. With 17 and 25 the console still reads The sum is 42, because 42 fits in the low byte the out writes. Type 200 and 100 instead: the 16 bit port prints 300 and the byte port prints 44, which is 300 with everything above eight bits thrown away.