How can I detect integer overflow on 32 bits int?

integer overflow c++
how to check 32 bit signed integer in java
integer overflow in c
integer overflow calculator
check integer overflow java
how to handle integer overflow in c
what is integer overflow how to avoid it
how to detect overflow

I know such topic was asked several times, but my question is about overflow on full 32 bits of int. For example:

  11111111111111111111111111111111 +
  00000000000000000000000000000001 =
  00000000000000000000000000000000   //overflow!

I found topic with similar question about this, however the algorithm is not perfect.

  11111111111111111111111111111111 +
  00000000000000000000000000000000 =
  00000000000000000000000000000000  //overflow!

Is there any simple, fast, safer way to check this ?

Check for Integer Overflow, Write a “C” function, int addOvf(int* result, int a, int b) If there is no overflow, the function places the resultant = sum a+b in The solution of casting to long and adding to find detecting the overflow is not allowed. #include <bits/stdc++.h>. Since we know the boundary values of integer, we can use them as a reference to detect integer overflow in C++. #include<bits/stdc++.h> using namespace std; /* Check if adding x and y results in overflow.

long test = (long)x+y;
if (test > Integer.MAX_VALUE || test < Integer.MIN_VALUE)
   // Overflow!

Catching Integer Overflows in C, Integers have finite ranges in computers, for example a 32-bit unsigned there is no value than can hold 2147483648, so if you negate (int)0x80000000, you get the easiest way to find out if that overflows is to check whether a+b<a (or b;  Check for Integer Overflow. Write a “C” function, int addOvf(int* result, int a, int b) If there is no overflow, the function places the resultant = sum a+b in “result” and returns 0. Otherwise it returns -1. The solution of casting to long and adding to find detecting the overflow is not allowed.

Try this way:

boolean isOverflow(int left, int right) {
    return right > 0
            ? Integer.MAX_VALUE - right < left
            : Integer.MIN_VALUE - right > left;
}

From: https://wiki.sei.cmu.edu/confluence/display/java/NUM00-J.+Detect+or+prevent+integer+overflow

Integer overflow, This integer types have fixed size; usually 8-bit, 16-bit, 32-bit, or 64-bit. -3+1 raised CONSTRAINT_ERROR : overflow.adb:49 range check failed +​2147483647 3 print (1+max int) 1 a68g: runtime error: 1: INT math error  An integer of type int in Java can be negative or positive, which means with its 32 bits, we can assign values between -2 31 (-2147483648) and 2 31-1 (2147483647). The wrapper class Integer defines two constants that hold these values: Integer.MIN_VALUE and Integer.MAX_VALUE .

Overflow can be detected by a logical expression of the most significant bit of the two operands and the (truncated) result (I took the logical expression from the MC68030 manual):

/**
 * Add two int's with overflow detection (r = s + d)
 */
public static int add(int s, int d) throws ArithmeticException {
    int r = s + d;
    if (((s & d & ~r) | (~s & ~d & r)) < 0)
        throw new ArithmeticException("int overflow add(" + s + ", " + d + ")");
    return r;
}

Integer overflow, int. 32-bit signed two's-complement. −2,147,483,648 to 2,147,483,647. long. 64-​bit This compliant solution uses the BigInteger technique to detect overflow:. i might be on a 32 bit actually. go -v says 'go version go1.3.1 darwin/386'. Is it possible to make this work on a 32 bit system? – sourcey Oct 3 '14 at 19:59 @sourcey can you modify your code to use int64 explicitly? if so - you're good to go.

The most intuitive method I can think of: calculate the sum (or difference) as a long, then convert that sum to an int and see if its value has changed.

long longSum = (long) a + b;
int sum = (int) longSum;
if (sum == longSum) {
    // sum contains the correct result
} else {
    // overflow/underflow
}

Remember that on modern 64 bit processors, working with longs is no less efficient than working with ints (the opposite may be true). So if you have a choice between checking for overflows or using longs, go for the latter.

NUM00-J. Detect or prevent integer overflow, To check for Integer overflow, we need to check the Integer.MAX_VALUE, which is the maximum value of an integer in Java. public class Demo { public static void main(String[] args) { int val1 = 9898989; int val2 = 6789054;  \$\begingroup\$ If you just change from using int to using unsigned int, or better still, uint32_t and size_t, you'll be able to do those checks after the operation.For signed ints, overflow and underflow can't be detected after-the-fact because of undefined behaviour.

Java Program to check for Integer overflow, Also see How to detect integer overflow in C/C++? on Stack Overflow. In two complements one can be pedantic and just pick out the sign bit to do operations  I know such topic was asked several times, but my question is about overflow on full 32 bits of int. For example: 11111111111111111111111111111111

Simple method to detect int overflow, 2: Integer overflows 2.1 Widthness overflows 2.1.1 Exploiting 2.2 Arithmetic overflows What happens is that the variable s is promoted to an int (32 bits long​), then the it is possible to bypass the bounds check at [w1] and overflow the buffer. Integer overflow can be demonstrated through an odometer overflowing, a mechanical version of the phenomenon. All digits are set to the maximum 9 and the next increment of the white digit causes a cascade of carry-over additions setting all digits to 0, but there is no higher digit (1,000,000s digit) to change to a 1, so the counter resets to zero.

Basic Integer Overflows, 4-bit two's complement representation. Signed Integer. Unsigned Integer. Standard Integer represent a value. Overflow Examples 1. ○. 1. int i;. ○. 2. unsigned int j;. ○. 3. i = INT_MAX; // 64-bit integer. // and check against 32-bit UINT_MAX. [The Joy of Programming] Integer overflows often result in nasty bugs. In this column, we’ll look at some techniques to detect an overflow before it occurs.

Comments
  • securecoding.cert.org/confluence/display/java/…
  • BTW, Java int is always 32 bits (regardless of the architecture that your given JVM is running on [stackoverflow.com/questions/18017752/…): "By default, the int data type is a 32-bit signed two's complement integer, which has a minimum value of -231 and a maximum value of 231-1" (docs.oracle.com/javase/tutorial/java/nutsandbolts/…). So the 32-bit specification here seems to be a bit irrelevant/moot.
  • Are x & y assumed to be long's themselves? If they were instead int's, I don't think this would work (since the overflow would go undetected before the long cast); am I correct?
  • You are not correct. X gets cast to long before addition (precedence). long+int the int gets converted to long and you end up with long+long. Therefore the entire operation is done at 64 bit precision.
  • Ah ok so the cast has precedence over the addition operator. Thanks!
  • Overflow in isOverflow(-47483648, -100000000) is not detected.
  • This is not an overflow actually :)
  • Java 8's addExact is coded as if (((s ^ r) & (d ^ r)) < 0).