Arithmetic Overflow in mips

Question

I am just started learning exception handler of MIPS instruction.

I need to make my program to have Arithmetic overflow exception so that i can test my exception handler.

I have two array A and B. Array A has hex number and Array B has integers.

How to make overflow by adding hex number and integer ?

The addition of which hex number and integer can cause overflow?

Solution

According to the MIPS instruction reference, the only addition operations which can produce overflow exceptions are the signed addition instructions:

ADD
ADDI

MIPS integers are 32-bit, and since you'll be using signed integers, the maximum value is 2³¹-1 (aka 2147483647 or hex 7FFFFFFF). Thus any addition which results in a number larger than this should throw an exception, e.g if you try to add 1 to 2147483647:

# Load 2147483647 into $s1
LUI $s0, 32767
ORI $s1, $s0, 65535

# Add 1 to $s1 and store in $s2. This should produce an overflow exception
ADDI $s2, $s1, 1

Comments

[Simon MᶜKenzie]

I'm not familiar with MIPS, but hex is just a representation of a number, not a datatype, so surely you're just adding two integers. If your integers are 32 bits, then any two numbers which add to more than 2^32-1 (i.e. 4294967295) will produce an overflow.

[Dennis]

Thank you Simon, i tried it already. I tried like this 0x1+4294967295. but I still don't get an overflow error. i get the answer 0x0.

[Simon MᶜKenzie]

Are you using add or addi? As far as I can see, if you use signed instructions, you should get overflow exceptions.

[Dennis]

i used "add". i should get overflow right?

[Simon MᶜKenzie]

The problem is probably that if you're using signed operations, the largest (signed) integer is actually 2^31-1. Try adding 1 to that.

[Dennis]

Wow !!! It works. I got the overflow exception error. It look like i am the only one who is happy to get the error. LOL Thanks a lot Simon. You are a great help. :D

[Roronoa Zoro]

How come is the maximum value (2^31) -1? Your range for an unsigned number is from 0 to (2^32) -1, using 32-bit integers!

[Simon MᶜKenzie]

Very good point @RoronoaZoro - I'm not sure why I wrote "unsigned" - ADD and ADDI are signed operations (which I even mentioned at the top of the answer). I've removed the "un"!