## How can I detect integer overflow on 32 bits int?

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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 `long`

s is no less efficient than working with `int`

s (the opposite may be true). So if you have a choice between checking for overflows or using `long`

s, 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)`

.