CODEWITHSUNDEEP
mipsunsigned
dankilman
dankilman

Reputation: 886

Why do MIPS operations on unsigned numbers give signed results?

When I try working on unsigned integers in MIPS, the result of every operation I do remains signed (that is, the integers are all in 2's complement), even though every operation I perform is an unsigned one: addu, multu and so fourth...

When I print numbers in the range [2^31, 2^32 - 1] I get their 'overflowed' negative value as if they were signed (I guess they are).

Though, when I try something like this:

li $v0, 1
li $a0, 2147483648                # or any bigger number
syscall

the printed number is always 2147483647 (2^31 - 1)

I'm confused... What am I missing?

PS : I haven't included my code as it isn't very readable (such is assembly code) and putting aside this problem, seems to be working fine. If anyone feels it is necessary I shall include it right away!

Upvotes: 15

Views: 15287

Answers (2)

John Knoeller
John Knoeller

Reputation: 34148

It looks to me like the real problem is the syscall that you are using to print numbers. It appears to and to always interpret what you pass as signed, and to possibly to bound as as well.

Upvotes: 2

Pedro Silva
Pedro Silva

Reputation: 4700

From Wikipedia:

The MIPS32 Instruction Set states that the word unsigned as part of Add and Subtract instructions, is a misnomer. The difference between signed and unsigned versions of commands is not a sign extension (or lack thereof) of the operands, but controls whether a trap is executed on overflow (e.g. Add) or an overflow is ignored (Add unsigned). An immediate operand CONST to these instructions is always sign-extended.

From the MIPS Instruction Reference:

ALL arithmetic immediate values are sign-extended [...] The only difference between signed and unsigned instructions is that signed instructions can generate an overflow exception and unsigned instructions can not.

Upvotes: 23

Related Questions