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I have searched a bit and didn't find anything that properly explained this.

In MIPS you have add and addu to do additions. The main difference is that addu doesn't generate an overflow exceptions.

Let's say we have this binary (I'm using four bits although MIPS is 32 for simplification):

0111

If we add 1 it becomes 1000.

With the add instruction there is an overflow and no carry, since a positive 7 became a negative 1 (assuming two's complement in MIPS). This generates an overflow exception too.

With the addu there is no overflow and no carry, since everything went as expected.

Now let's say you have this binary:

1111

If we add 1 it becomes 0000.

With the add instruction there should be no overflow since a negative 1 became a 0. What about the carry out flag, does it change to 1?

And what happens with the addu instruction? Is it considered overflow, since 15 became 0? I know that there is no overflow exception, but what happens to the overflow flag? Does it get set to 1? What about the carry flag?

There is no carryout flag in MIPS architecture.

Overflow exception in the case of addition happens when the sign of the result is different form the expected sign which can happen:

• When you add two positive numbers and you get a negative number as a result


• When you add two negative numbers and you get a positive number as a result

So yes, in you hypothetical 4 bit MIPS 0111+1 results in an overflow, with the corresponding flag being activated.