Bitwise overflow checking in c [closed]

Question

Closed. This question needs debugging details. It is not currently accepting answers.

---

Want to improve this question? Update the question so it's on-topic for Stack Overflow.

Closed 5 years ago.

Improve this question

I am trying to write two functions that will check/prevent overflow in c (using only ! ~ | & ^ +) but cant get it. The first is will a certain twos compliment/signed int will fit in a certatin amount of bits: fitsB(int x, int n) where is the int and n is the size of bits to use. Also a function that will check to see if two ints will not overflow when added together: overflowInt(int x, int y). I can get it if they are unsigned ints but the negatives just make things harder for me. Anyone know how to?

There also is no casting and ints are always 32 bit

Answer 1

/* 
 * addOK - Determine if can compute x+y without overflow
 *   Example: addOK(0x80000000,0x80000000) = 0,
 *            addOK(0x80000000,0x70000000) = 1, 
 *   Legal ops: ! ~ & ^ | + << >>
 *   Max ops: 20
 *   Rating: 3
 */
 int addOK(int x, int y) {
  // Find the sign bit in each word
  //if a and b have different signs, you cannot get overflow.
  //if they are the same, check that a is different from c and b is different from c, 
  // if they are the same, then there was no overflow.
  int z=x+y;
  int a=x>>31;
  int b=y>>31;
  int c=z>>31;
  return !!(a^b)|(!(a^c)&!(b^c));
}

Answer 2

x will fit in n bits if x < 2^(n-1).

The overflow question needs more information. Two ints will not overflow if you assign them to a long (or a double).

Answer 3

Using the above example (Adam Shiemke), you can find the maximum (positive) value and minimum value (negative) to get the range for n number of bits. 2^(n-1) (from Adam's example) and minus one for the maximum/positive number which can be represented in the n bits. For the minimum value, negate 2^(n-1) to get the minimum value x => -(2^(n-1)); (Note the >= not > for the minimum range). For example, for n = 4 bits, 2^(4-1) - 1 = 2^3 -1 = 7 so x <= 7 and x >= -8 = (-(2^(4-1)).

This assumes the initial input does not overflow a 32 bit quanity (Hopefully an error occurs in that condition) and the number of bits you are using is less then 32 (as you are adding 1 for the negative range and if you have 32 bits, it will overflow, see below for an explanation).

To determine if addition will overflow, if you have the maximum value, the x + y <= maximum value. By using algebra, we can get y <= maximum value - x. You can then compare the passed in value for y and if it does not meet the condition, the addition will overflow. For example if x is the maximumn value, then y <= 0, so y must be less then or equal to zero or the addition will overflow.