Arithmetic Overflow and Underflowing 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?
 A: 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.
A: 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.
A: 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...)  
