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One of my lecture slides gives an example of arithmetic overflow and carry in a topic for conditional branching flags on an ARM chip, quoted below:
Presumably for the sake of the example, the address can only hold 8 bytes. So to me, it seems likes adding 1 to 7FFFFFFF gives 80000000. I thought 80000000 would still fit into an 8-byte address.
Why is this an arithmetic overflow? Is it the wrong way around on the slide? Or is my understanding flawed?
Thanks for any responses
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Carry indicates the result isn't mathematically correct when interpreted as unsigned, overflow indicates the result isn't mathematically correct when interpreted as signed. So as examples for your 4-bit ALU: 1111 + 0001 = 0000 should set carry (15 + 1 = 0 is false) and clear overflow (-1 + 1 = 0 is true).
From a mechanistic point of view, the carry flag is set when there is a carry out of the most-significant bit. The overflow flag is set when the carry into the most significant bit is different from the carry out of it. With unsigned arithmetic you only have to worry about the carry flag.
An arithmetic overflow is the result of a calculation that exceeds the memory space designated to hold it. For example, a divide-by-zero yields a much larger result. See arithmetic underflow.
So when adding binary numbers, a carry out is generated when the “SUM” equals or is greater than two (1+1) and this becomes a “CARRY” bit for any subsequent addition being passed over to the next column for addition and so on.
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