MIPS – why does the following print “OK”?

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The following fragment of MIPS code results in “OK” being printed when I run it on QtSpim.

But why is this?

addi	$v0,	$zero,	11
addi	$a0,	$zero,	79
syscall
addi	$v0,	$zero,	11
addi	$a0,	$zero,	75
syscall
addi	$v0,	$zero,	10
syscall

What I have tried:

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

But they only store the values temporarily, and the end result is "OK", but I don't know why.

Posted 3-Dec-18 12:16pm

Kaiser89

Updated 3-Dec-18 21:22pm

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CPallini · Accepted Answer · 2018-12-03T21:23:00

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

Your assumptions are wrong.

ASM

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

ASM

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.