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I am a bit unclear about underflowing in terms of binary representation.

Let's say that an unsigned 8-bit variable gets overflown from the addition of $150+150$.

A signed 8-bit variable gets underflown after the subtraction of $-120-60$.

Now my point is let's think of 8-bit variable, we are subtracting $110-10$. Now let's convert this into an addition, $110+(-10)$. Since $-10$ is $11110110$ and $110$ is $01101110$. If we add these two binary numbers we will have a value after 8th bit to carry, which is I believe an overflown, however the final binary number is equal to $100$ and that's what we want and in terms of decimal value we did not lose anything. In that case do we have a overflow or underflow here?

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    $\begingroup$ Underflow makes no sense for integral values. $\endgroup$ – copper.hat Oct 21 '13 at 16:36
  • $\begingroup$ @copper.hat That's not quite true: cwe.mitre.org/data/definitions/191.html $\endgroup$ – lehins Jan 31 at 14:23
  • $\begingroup$ That is overflow, not underflow. $\endgroup$ – copper.hat Jan 31 at 14:48
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Say you have $8$-bits signed integers. The range of representable integers start at $-128$ and ends at $127$.

If you perform $127+1$, you obtain $-128$ : $0111 1111+0000 0001 = 1000 0000$ and the overflow flag is turned on.

If you perform $-128-1$, you obtain $127$ : $1000 0000-0000 0001 = 0111 1111$ and the overflow flag is turned on.

In both cases, these are overflows.

Underflows refer to floating point underflow, where an operation result in a number that is too small to be representable. For example, if the exponent part can represent values from $-127$ to $127$, then any number with absolute value less than $2^{-127}$ may cause underflow.

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You don't have an overflow here: the result will be 01100100. Since the top bit indicates the sign, the addition process is not the same as for unsigned integers.

For example, 01100100+01100100 overflows for signed integers, because we can't carry from the 7th bit into 8th: the 8th bit is the sign. Over unsigned integers, there is no overflow.

Conversely, 01100100+11110110 overflows for unsigned integers, but not for signed ones.

My point is: the process of adding of numbers represented by binary strings depends on what representation is used. (You would not add floating point representations bitwise...)

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The term arithmetic underflow (or "floating point underflow", or just "underflow") is a condition in a computer program where the result of a calculation is a number of smaller absolute value than the computer can actually store in memory.

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