What is the diffrence between a carry bit and overflow
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$\begingroup$ What have you tried so far? The title contains different question from body. Reading flags is strictly programming task, so off-topic here. $\endgroup$– EvilFeb 10 '18 at 16:42
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$\begingroup$ en.wikipedia.org/wiki/Status_register $\endgroup$– HEKTOFeb 10 '18 at 17:01
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$\begingroup$ This video solved my problem. Hopefully it does for you too: youtu.be/9cXe_T99nL4 $\endgroup$– a-kihliFeb 10 '18 at 17:03
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$\begingroup$ you should mention the architecture you are referring to. $\endgroup$– Ran G.Feb 10 '18 at 18:36
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$\begingroup$ @RanG. If the answer depends on any particular architecture, it's hard to see it being on-topic here. $\endgroup$– David RicherbyFeb 11 '18 at 13:10
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Carry is generally used for unsigned arithmetic and overflow is used for signed arithmetic.
This unsigned 8-bit operation results in Carry, but no overflow (the sign of the result is correct):
0xC0 + 0xD8 = 0x98
If we're doing unsigned 8-bit arithmetic, that's fine and we're only interested in the carry bit in this case, which tells us that the "correct" answer is actually 0x198.
If we look at the same operation and the same operand values, but consider that the operands are signed, we have the equivalent:
-0x40 + -0x28 = -0x68
In this case, there is also no "overflow" and the result of the arithmetic is correct.
Consider this unsigned operation, however:
0x70 + 0x68 = 0xD8
No overflow here -- and no carry. The unsigned arithmetic is simple. But now look at the operands as signed values:
0x70 + -0x98 = 0x28
The answer is obviously wrong. We have subtracted a larger number from a smaller one and ended up with a positive value. The carry is not set here, so it doesn't offer a clue of the problem. The overflow flag reveals that the "correct" signed answer is -0x28.