Carry vs Overflow in Math

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.

See also:

• http://www.geocities.com/SiliconValley/2499/essays.html#TWOS

Comments:

• The reason the last example is incorrect is because you're interpreting it as a negative overflow. 0x68 is a positive number if it's a signed 8 bits. The carry bit is set because 0x68 + 0x70 is 216, which is certainly greater than the largest positive 8 bit signed number (127). The result will be interpreted as a negative number, namely -128 + 88 = -40. The result we want however is +128 + 88 = 216.+
• your last example is obviosly uncorrect James Newton of MassMind replies: yes... that was the point of the example...+

  +

• I believe the last example is incorrect. Adding a positive and negative number *never* results in overflow. The overflow flag is not the same as the sign bit. Overflow occurs when two large negative numbers are added and result in a positive, or when two large positive numbers are added and result in a negative.+

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