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 syscall.

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 "syscall" lecture. What is new here is a service that answers, service 5 leaves the number that was typed in $v0, which is the same register the next service number goes into.

$v0 does two jobs on this machine and they collide here. It is where the service number goes, and it is where a service that answers puts its answer, so the number service 5 read is in $v0 for exactly as long as it takes you to write the next li $v0. move $t0, $v0 straight after the read is the habit that keeps it, and the second read adds $v0 into $t0 before anything overwrites it.

The M68K asks for the prompt and the number in one request, task 18. Here they are two services, a 4 and a 5, so the prompt has to be printed before the read every time.

The \n at the front of second and answer is a newline inside the string, which is how you get a line break without a service of its own. .asciiz "\nSecond number: " is one string of seventeen bytes and the first of them is the line break.

Type 17 and 25 and the console reads The sum is 42. Service 5 reads a decimal number and nothing else, and a line that is not one ends the run.

Try changing add $t0, $t0, $v0 to sub $t0, $t0, $v0. With 17 and 25 the console reads The sum is -8, because service 1 prints its argument as a signed 32 bit number: the bits FFFFFFF8 are -8 to it. Change the li $v0, 1 below it to li $v0, 36 and the same bits print as 4294967288.