Check whether an integer, or an addition, subtraction or multiplication result, overflows a signed or unsigned 8-, 16-, 32- or 64-bit type.
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Overflow Error Formula
The following formula measures only the amount by which an addition exceeds an upper bound. It is not the wrapped result and does not detect values below a minimum. Compare the exact result with both bounds for a complete range check.
Variables:
- E is the amount above the maximum, or zero if the sum does not exceed it
- CV is the current value
- I is the increment value
- MV is the maximum value that can be held without overflow
For an n-bit unsigned integer, the range is 0 to 2ⁿ − 1. For signed two’s-complement, it is −2ⁿ⁻¹ to 2ⁿ⁻¹ − 1. Calculate exactly, then compare with both bounds. Illustrative wrapping takes the result modulo 2ⁿ; for signed display, subtract 2ⁿ when the normalized value is at least 2ⁿ⁻¹.
What is an Overflow Error?
Integer overflow occurs when an exact arithmetic result lies outside the selected integer type’s representable range, either above its maximum or below its minimum. A programming language may wrap, report an error or specify other behavior. This calculator shows a mathematical range check and an illustrative modulo result; it does not simulate a particular language’s operand promotion or checked-arithmetic rules.
How to Calculate Overflow Error?
The following steps outline how to calculate the Overflow Error.
- First, choose the integer type and bit width to determine its minimum and maximum (MV).
- Next, determine the current value (CV) before the increment.
- Next, determine the increment value (I) that will be added to the current value.
- Next, for this addition example, compute E = max(0, (CV + I) − MV).
- Finally, E greater than zero means the sum exceeds the maximum. Also compare the exact sum with the minimum; E alone does not detect a result below that bound.
- After inserting the variables and calculating the result, check your answer with the calculator above.
Example Problem :
Use the following variables as an example problem to test your knowledge.
maximum value (MV) = 255
current value (CV) = 250
increment value (I) = 10
The exact sum is 250 + 10 = 260. For an 8-bit unsigned type, the maximum is 255, so the amount above maximum is 5. The illustrative modular result is 260 mod 256 = 4. These are different quantities.
