MIPS – why does the following print “OK”?

Quote:

I expected the register $v0 to store 32 (11 + 11 + 10) and $a0 to store 154 (79 + 75).

Your assumptions are wrong.

addi $v0, $zero, 11

could be written (pseudocode)

v0 <- ([$zero]+11)

That is: add the immediate integer 11 with the content of register 0 and store the result into register v0.


So the sequence

addi	$v0,	$zero,	11
addi	$a0,	$zero,	79
syscall

is

$v0 <- 11
$a0 <- 79
syscall

syscall is invoked with code 11 ('print_character') and argument 79 (ASCII code of 'O').
A similar argument applies to the remaining lines of the program.