**40**results.

If it overflows, it goes back to the minimum value and continues from there. If it underflows, it goes back to the maximum value and continues from there.

You can check that beforehand as follows:

int

long

If you think that this may occur more than often, then consider using a datatype or object which can store larger values, e.g. long or maybe java.math.BigInteger. The last one doesn't overflow, practically, the available JVM memory is the limit.

If you happen to be on Java8 already, then you can make use of the new Math#addExact() and Math#subtractExact() methods which will throw an ArithmeticException on overflow.

public static boolean willAdditionOverflow(int left, int right) { try { Math.addExact(left, right); return false; } catch (ArithmeticException e) { return true; } } public static boolean willSubtractionOverflow(int left, int right) { try { Math.subtractExact(left, right); return false; } catch (ArithmeticException e) { return true; } }

The source code can be found here and here respectively.

Of course, you could also just use them right away instead of hiding them in a boolean utility method.

If it overflows, it goes back to the minimum value and continues from there

continues from there

@dhblah, Let's say the maximum and minimum values that Java allows for an int are +100, -100, respectively. If you were adding one to a Java integer, the process would look like this as it overflowed. 98, 99, 100, -100, -99, -98, .... Does that make more sense?

It would be great if this answer could be updated to include the new Java 8 solution(s) in java.lang.Math

## How does Java handle integer underflows and overflows and how would yo...

public static void main(String[] args) { int i = Integer.MAX_VALUE; int j = Integer.MIN_VALUE; System.out.println(i+1); System.out.println(j-1); }

Well! And now, can you answer, how to detect it into complex calculus?

## How does Java handle integer underflows and overflows and how would yo...

By default, Java's int and long math silently wrap around on overflow and underflow. (Integer operations on other integer types are performed by first promoting the operands to int or long, per JLS 4.2.2.)

As of Java 8, java.lang.Math provides addExact, subtractExact, multiplyExact, incrementExact, decrementExact and negateExact static methods for both int and long arguments that perform the named operation, throwing ArithmeticException on overflow. (There's no divideExact method -- you'll have to check the one special case (MIN_VALUE / -1) yourself.)

As of Java 8, java.lang.Math also provides toIntExact to cast a long to an int, throwing ArithmeticException if the long's value does not fit in an int. This can be useful for e.g. computing the sum of ints using unchecked long math, then using toIntExact to cast to int at the end (but be careful not to let your sum overflow).

If you're still using an older version of Java, Google Guava provides IntMath and LongMath static methods for checked addition, subtraction, multiplication and exponentiation (throwing on overflow). These classes also provide methods to compute factorials and binomial coefficients that return MAX_VALUE on overflow (which is less convenient to check). Guava's primitive utility classes, Bytes, Shorts and Ints, provide checkedCast methods for narrowing larger types (throwing IllegalArgumentException on under/overflow, not ArithmeticException), as well as saturatingCast methods that return MIN_VALUE or MAX_VALUE on overflow.

## How does Java handle integer underflows and overflows and how would yo...

There are libraries that provide safe arithmetic operations, which check integer overflow/underflow . For example, Guava's IntMath.checkedAdd(int a, int b) returns the sum of a and b, provided it does not overflow, and throws ArithmeticException if a + b overflows in signed int arithmetic.

## How does Java handle integer underflows and overflows and how would yo...

public static void main(String[] args) { int i = Integer.MAX_VALUE; int j = Integer.MIN_VALUE; System.out.println(i+1); System.out.println(j-1); }

Well! And now, can you answer, how to detect it into complex calculus?

## How does Java handle integer underflows and overflows and how would yo...

There is one case, that is not mentioned above:

int res = 1; while (res != 0) { res *= 2; } System.out.println(res);

## How does Java handle integer underflows and overflows and how would yo...

It basically handles them without reporting an exception, performing the 2's complement arithmetic without concern for overflow or underflow, returning the expected (but incorrect) result based on the mechanics of 2's complement arithmetic.

This means that the bits which over or underflow are simply chopped, and that Integer.MIN_VALUE - 1 typically returns Integer.MAX_VALUE.

As far as "lower order bytes" being a workaround, they really aren't. What is happening when you use Java bytes to do the arithmetic is that they get expanded into ints, the arithmetic is generally performed on the ints, and the end result is likely to be completely contained in the returned it as it has far more storage capacity than the starting bytes.

## How does java handle integer overflow and underflow? - Stack Overflow

Well, as far as primitive integer types go, Java doesnt handle Over/Underflow at all (for float and double the behaviour is different, it will flush to +/- infinity just as IEEE-754 mandates).

When adding two int's, you will get no indication when an overflow occurs. A simple method to check for overflow is to use the next bigger type to actually perform the operation and check if the result is still in range for the source type:

public int addWithOverflowCheck(int a, int b) { // the cast of a is required, to make the + work with long precision, // if we just added (a + b) the addition would use int precision and // the result would be cast to long afterwards! long result = ((long) a) + b; if (result > Integer.MAX_VALUE) { throw new RuntimeException("Overflow occured"); } else if (result < Integer.MIN_VALUE) { throw new RuntimeException("Underflow occured"); } // at this point we can safely cast back to int, we checked before // that the value will be withing int's limits return (int) result; }

What you would do in place of the throw clauses, depends on your applications requirements (throw, flush to min/max or just log whatever). If you want to detect overflow on long operations, you're out of luck with primitives, use BigInteger instead.

Edit (2014-05-21): Since this question seems to be referred to quite frequently and I had to solve the same problem myself, its quite easy to evaluate the overflow condition by the same method a CPU would calculate its V flag.

Its basically a boolean expression that involves the sign of both operands as well as the result:

/** * Add two int's with overflow detection (r = s + d) */ public static int add(final int s, final 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; }

In java its simpler to apply the expression (in the if) to the entire 32 bits, and check the result using < 0 (this will effectively test the sign bit). The principle works exactly the same for all integer primitive types, changing all declarations in above method to long makes it work for long.

For smaller types, due to the implicit conversion to int (see the JLS for bitwise operations for details), instead of checking < 0, the check needs to mask the sign bit explicitly (0x8000 for short operands, 0x80 for byte operands, adjust casts and parameter declaration appropiately):

/** * Subtract two short's with overflow detection (r = d - s) */ public static short sub(final short d, final short s) throws ArithmeticException { int r = d - s; if ((((~s & d & ~r) | (s & ~d & r)) & 0x8000) != 0) throw new ArithmeticException("short overflow sub(" + s + ", " + d + ")"); return (short) r; }

(Note that above example uses the expression need for subtract overflow detection)

So how/why do these boolean expressions work? First, some logical thinking reveals that an overflow can only occur if the signs of both arguments are the same. Because, if one argument is negative and one positive, the result (of add) must be closer to zero, or in the extreme case one argument is zero, the same as the other argument. Since the arguments by themselves can't create an overflow condition, their sum can't create an overflow either.

So what happens if both arguments have the same sign? Lets take a look at the case both are positive: adding two arguments that create a sum larger than the types MAX_VALUE, will always yield a negative value, so an overflow occurs if arg1 + arg2 > MAX_VALUE. Now the maximum value that could result would be MAX_VALUE + MAX_VALUE (the extreme case both arguments are MAX_VALUE). For a byte (example) that would mean 127 + 127 = 254. Looking at the bit representations of all values that can result from adding two positive values, one finds that those that overflow (128 to 254) all have bit 7 set, while all that do not overflow (0 to 127) have bit 7 (topmost, sign) cleared. Thats exactly what the first (right) part of the expression checks:

if (((s & d & ~r) | (~s & ~d & r)) < 0)

(~s & ~d & r) becomes true, only if, both operands (s, d) are positive and the result (r) is negative (the expression works on all 32 bits, but the only bit we're interested in is the topmost (sign) bit, which is checked against by the < 0).

Now if both arguments are negative, their sum can never be closer to zero than any of the arguments, the sum must be closer to minus infinity. The most extreme value we can produce is MIN_VALUE + MIN_VALUE, which (again for byte example) shows that for any in range value (-1 to -128) the sign bit is set, while any possible overflowing value (-129 to -256) has the sign bit cleared. So the sign of the result again reveals the overflow condition. Thats what the left half (s & d & ~r) checks for the case where both arguments (s, d) are negative and a result that is positive. The logic is largely equivalent to the positive case; all bit patterns that can result from adding two negative values will have the sign bit cleared if and only if an underflow occured.

long

This will work but I'm assuming it will have a nasty performance hit.

## How does Java handle integer underflows and overflows and how would yo...

If it overflows, it goes back to the minimum value and continues from there. If it underflows, it goes back to the maximum value and continues from there.

You can check that beforehand as follows:

int

long

If you think that this may occur more than often, then consider using a datatype or object which can store larger values, e.g. long or maybe java.math.BigInteger. The last one doesn't overflow, practically, the available JVM memory is the limit.

If you happen to be on Java8 already, then you can make use of the new Math#addExact() and Math#subtractExact() methods which will throw an ArithmeticException on overflow.

public static boolean willAdditionOverflow(int left, int right) { try { Math.addExact(left, right); return false; } catch (ArithmeticException e) { return true; } } public static boolean willSubtractionOverflow(int left, int right) { try { Math.subtractExact(left, right); return false; } catch (ArithmeticException e) { return true; } }

The source code can be found here and here respectively.

Of course, you could also just use them right away instead of hiding them in a boolean utility method.

If it overflows, it goes back to the minimum value and continues from there

continues from there

@dhblah, Let's say the maximum and minimum values that Java allows for an int are +100, -100, respectively. If you were adding one to a Java integer, the process would look like this as it overflowed. 98, 99, 100, -100, -99, -98, .... Does that make more sense?

It would be great if this answer could be updated to include the new Java 8 solution(s) in java.lang.Math

## How does Java handle integer underflows and overflows and how would yo...

Java doesn't do anything with integer overflow for either int or long primitive types and ignores overflow with positive and negative integers.

This answer first describes the of integer overflow, gives an example of how it can happen, even with intermediate values in expression evaluation, and then gives links to resources that give detailed techniques for preventing and detecting integer overflow.

Integer arithmetic and expressions reslulting in unexpected or undetected overflow are a common programming error. Unexpected or undetected integer overflow is also a well-known exploitable security issue, especially as it affects array, stack and list objects.

Overflow can occur in either a positive or negative direction where the positive or negative value would be beyond the maximum or minimum values for the primitive type in question. Overflow can occur in an intermediate value during expression or operation evaluation and affect the outcome of an expression or operation where the final value would be expected to be within range.

Sometimes negative overflow is mistakenly called underflow. Underflow is what happens when a value would be closer to zero than the representation allows. Underflow occurs in integer arithmetic and is expected. Integer underflow happens when an integer evaluation would be between -1 and 0 or 0 and 1. What would be a fractional result truncates to 0. This is normal and expected with integer arithmetic and not considered an error. However, it can lead to code throwing an exception. One example is an "ArithmeticException: / by zero" exception if the result of integer underflow is used as a divisor in an expression.

Consider the following code:

which results in x being assigned 0 and the subsequent evaluation of bigValue / x throws an exception, "ArithmeticException: / by zero" (i.e. divide by zero), instead of y being assigned the value 2.

The expected result for x would be 858,993,458 which is less than the maximum int value of 2,147,483,647. However, the intermediate result from evaluating Integer.MAX_Value * 2, would be 4,294,967,294, which exceeds the maximum int value and is -2 in accordance with 2s complement integer representations. The subsequent evaluation of -2 / 5 evaluates to 0 which gets assigned to x.

Rearranging the expression for computing x to an expression that, when evaluated, divides before multiplying, the following code:

int bigValue = Integer.MAX_VALUE; int x = bigValue / 5 * 2; int y = bigValue / x;

results in x being assigned 858,993,458 and y being assigned 2, which is expected.

The intermediate result from bigValue / 5 is 429,496,729 which does not exceed the maximum value for an int. Subsequent evaluation of 429,496,729 * 2 doesn't exceed the maximum value for an int and the expected result gets assigned to x. The evaluation for y then does not divide by zero. The evaluations for x and y work as expected.

Java integer values are stored as and behave in accordance with 2s complement signed integer representations. When a resulting value would be larger or smaller than the maximum or minimum integer values, a 2's complement integer value results instead. In situations not expressly designed to use 2s complement behavior, which is most ordinary integer arithmetic situations, the resulting 2s complement value will cause a programming logic or computation error as was shown in the example above. An excellent Wikipedia article describes 2s compliment binary integers here: Two's complement - Wikipedia

There are techniques for avoiding unintentional integer overflow. Techinques may be categorized as using pre-condition testing, upcasting and BigInteger.

Pre-condition testing comprises examining the values going into an arithmetic operation or expression to ensure that an overflow won't occur with those values. Programming and design will need to create testing that ensures input values won't cause overflow and then determine what to do if input values occur that will cause overflow.

Upcasting comprises using a larger primitive type to perform the arithmetic operation or expression and then determining if the resulting value is beyond the maximum or minimum values for an integer. Even with upcasting, it is still possible that the value or some intermediate value in an operation or expression will be beyond the maximum or minimum values for the upcast type and cause overflow, which will also not be detected and will cause unexpected and undesired results. Through analysis or pre-conditions, it may be possible to prevent overflow with upcasting when prevention without upcasting is not possible or practical. If the integers in question are already long primitive types, then upcasting is not possible with primitive types in Java.

The BigInteger technique comprises using BigInteger for the arithmetic operation or expression using library methods that use BigInteger. BigInteger does not overflow. It will use all available memory, if necessary. Its arithmetic methods are normally only slightly less efficient than integer operations. It is still possible that a result using BigInteger may be beyond the maximum or minimum values for an integer, however, overflow will not occur in the arithmetic leading to the result. Programming and design will still need to determine what to do if a BigInteger result is beyond the maximum or minimum values for the desired primitive result type, e.g., int or long.

The Carnegie Mellon Software Engineering Institute's CERT program and Oracle have created a set of standards for secure Java programming. Included in the standards are techniques for preventing and detecting integer overflow. The standard is published as a freely accessible online resource here: The CERT Oracle Secure Coding Standard for Java

The standard's section that describes and contains practical examples of coding techniques for preventing or detecting integer overflow is here: NUM00-J. Detect or prevent integer overflow

Book form and PDF form of The CERT Oracle Secure Coding Standard for Java are also available.

## How does Java handle integer underflows and overflows and how would yo...

public static void main(String[] args) { int i = Integer.MAX_VALUE; int j = Integer.MIN_VALUE; System.out.println(i+1); System.out.println(j-1); }

Well! And now, can you answer, how to detect it into complex calculus?

## How does Java handle integer underflows and overflows and how would yo...

There is one case, that is not mentioned above:

int res = 1; while (res != 0) { res *= 2; } System.out.println(res);

## How does Java handle integer underflows and overflows and how would yo...

Well, as far as primitive integer types go, Java doesnt handle Over/Underflow at all (for float and double the behaviour is different, it will flush to +/- infinity just as IEEE-754 mandates).

When adding two int's, you will get no indication when an overflow occurs. A simple method to check for overflow is to use the next bigger type to actually perform the operation and check if the result is still in range for the source type:

public int addWithOverflowCheck(int a, int b) { // the cast of a is required, to make the + work with long precision, // if we just added (a + b) the addition would use int precision and // the result would be cast to long afterwards! long result = ((long) a) + b; if (result > Integer.MAX_VALUE) { throw new RuntimeException("Overflow occured"); } else if (result < Integer.MIN_VALUE) { throw new RuntimeException("Underflow occured"); } // at this point we can safely cast back to int, we checked before // that the value will be withing int's limits return (int) result; }

What you would do in place of the throw clauses, depends on your applications requirements (throw, flush to min/max or just log whatever). If you want to detect overflow on long operations, you're out of luck with primitives, use BigInteger instead.

Edit (2014-05-21): Since this question seems to be referred to quite frequently and I had to solve the same problem myself, its quite easy to evaluate the overflow condition by the same method a CPU would calculate its V flag.

Its basically a boolean expression that involves the sign of both operands as well as the result:

/** * Add two int's with overflow detection (r = s + d) */ public static int add(final int s, final 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; }

In java its simpler to apply the expression (in the if) to the entire 32 bits, and check the result using < 0 (this will effectively test the sign bit). The principle works exactly the same for all integer primitive types, changing all declarations in above method to long makes it work for long.

For smaller types, due to the implicit conversion to int (see the JLS for bitwise operations for details), instead of checking < 0, the check needs to mask the sign bit explicitly (0x8000 for short operands, 0x80 for byte operands, adjust casts and parameter declaration appropiately):

/** * Subtract two short's with overflow detection (r = d - s) */ public static short sub(final short d, final short s) throws ArithmeticException { int r = d - s; if ((((~s & d & ~r) | (s & ~d & r)) & 0x8000) != 0) throw new ArithmeticException("short overflow sub(" + s + ", " + d + ")"); return (short) r; }

(Note that above example uses the expression need for subtract overflow detection)

So how/why do these boolean expressions work? First, some logical thinking reveals that an overflow can only occur if the signs of both arguments are the same. Because, if one argument is negative and one positive, the result (of add) must be closer to zero, or in the extreme case one argument is zero, the same as the other argument. Since the arguments by themselves can't create an overflow condition, their sum can't create an overflow either.

So what happens if both arguments have the same sign? Lets take a look at the case both are positive: adding two arguments that create a sum larger than the types MAX_VALUE, will always yield a negative value, so an overflow occurs if arg1 + arg2 > MAX_VALUE. Now the maximum value that could result would be MAX_VALUE + MAX_VALUE (the extreme case both arguments are MAX_VALUE). For a byte (example) that would mean 127 + 127 = 254. Looking at the bit representations of all values that can result from adding two positive values, one finds that those that overflow (128 to 254) all have bit 7 set, while all that do not overflow (0 to 127) have bit 7 (topmost, sign) cleared. Thats exactly what the first (right) part of the expression checks:

if (((s & d & ~r) | (~s & ~d & r)) < 0)

(~s & ~d & r) becomes true, only if, both operands (s, d) are positive and the result (r) is negative (the expression works on all 32 bits, but the only bit we're interested in is the topmost (sign) bit, which is checked against by the < 0).

Now if both arguments are negative, their sum can never be closer to zero than any of the arguments, the sum must be closer to minus infinity. The most extreme value we can produce is MIN_VALUE + MIN_VALUE, which (again for byte example) shows that for any in range value (-1 to -128) the sign bit is set, while any possible overflowing value (-129 to -256) has the sign bit cleared. So the sign of the result again reveals the overflow condition. Thats what the left half (s & d & ~r) checks for the case where both arguments (s, d) are negative and a result that is positive. The logic is largely equivalent to the positive case; all bit patterns that can result from adding two negative values will have the sign bit cleared if and only if an underflow occured.

long

This will work but I'm assuming it will have a nasty performance hit.

## How does Java handle integer underflows and overflows and how would yo...

Java doesn't do anything with integer overflow for either int or long primitive types and ignores overflow with positive and negative integers.

This answer first describes the of integer overflow, gives an example of how it can happen, even with intermediate values in expression evaluation, and then gives links to resources that give detailed techniques for preventing and detecting integer overflow.

Integer arithmetic and expressions reslulting in unexpected or undetected overflow are a common programming error. Unexpected or undetected integer overflow is also a well-known exploitable security issue, especially as it affects array, stack and list objects.

Overflow can occur in either a positive or negative direction where the positive or negative value would be beyond the maximum or minimum values for the primitive type in question. Overflow can occur in an intermediate value during expression or operation evaluation and affect the outcome of an expression or operation where the final value would be expected to be within range.

Sometimes negative overflow is mistakenly called underflow. Underflow is what happens when a value would be closer to zero than the representation allows. Underflow occurs in integer arithmetic and is expected. Integer underflow happens when an integer evaluation would be between -1 and 0 or 0 and 1. What would be a fractional result truncates to 0. This is normal and expected with integer arithmetic and not considered an error. However, it can lead to code throwing an exception. One example is an "ArithmeticException: / by zero" exception if the result of integer underflow is used as a divisor in an expression.

Consider the following code:

which results in x being assigned 0 and the subsequent evaluation of bigValue / x throws an exception, "ArithmeticException: / by zero" (i.e. divide by zero), instead of y being assigned the value 2.

The expected result for x would be 858,993,458 which is less than the maximum int value of 2,147,483,647. However, the intermediate result from evaluating Integer.MAX_Value * 2, would be 4,294,967,294, which exceeds the maximum int value and is -2 in accordance with 2s complement integer representations. The subsequent evaluation of -2 / 5 evaluates to 0 which gets assigned to x.

Rearranging the expression for computing x to an expression that, when evaluated, divides before multiplying, the following code:

int bigValue = Integer.MAX_VALUE; int x = bigValue / 5 * 2; int y = bigValue / x;

results in x being assigned 858,993,458 and y being assigned 2, which is expected.

The intermediate result from bigValue / 5 is 429,496,729 which does not exceed the maximum value for an int. Subsequent evaluation of 429,496,729 * 2 doesn't exceed the maximum value for an int and the expected result gets assigned to x. The evaluation for y then does not divide by zero. The evaluations for x and y work as expected.

Java integer values are stored as and behave in accordance with 2s complement signed integer representations. When a resulting value would be larger or smaller than the maximum or minimum integer values, a 2's complement integer value results instead. In situations not expressly designed to use 2s complement behavior, which is most ordinary integer arithmetic situations, the resulting 2s complement value will cause a programming logic or computation error as was shown in the example above. An excellent Wikipedia article describes 2s compliment binary integers here: Two's complement - Wikipedia

There are techniques for avoiding unintentional integer overflow. Techinques may be categorized as using pre-condition testing, upcasting and BigInteger.

Pre-condition testing comprises examining the values going into an arithmetic operation or expression to ensure that an overflow won't occur with those values. Programming and design will need to create testing that ensures input values won't cause overflow and then determine what to do if input values occur that will cause overflow.

Upcasting comprises using a larger primitive type to perform the arithmetic operation or expression and then determining if the resulting value is beyond the maximum or minimum values for an integer. Even with upcasting, it is still possible that the value or some intermediate value in an operation or expression will be beyond the maximum or minimum values for the upcast type and cause overflow, which will also not be detected and will cause unexpected and undesired results. Through analysis or pre-conditions, it may be possible to prevent overflow with upcasting when prevention without upcasting is not possible or practical. If the integers in question are already long primitive types, then upcasting is not possible with primitive types in Java.

The BigInteger technique comprises using BigInteger for the arithmetic operation or expression using library methods that use BigInteger. BigInteger does not overflow. It will use all available memory, if necessary. Its arithmetic methods are normally only slightly less efficient than integer operations. It is still possible that a result using BigInteger may be beyond the maximum or minimum values for an integer, however, overflow will not occur in the arithmetic leading to the result. Programming and design will still need to determine what to do if a BigInteger result is beyond the maximum or minimum values for the desired primitive result type, e.g., int or long.

The Carnegie Mellon Software Engineering Institute's CERT program and Oracle have created a set of standards for secure Java programming. Included in the standards are techniques for preventing and detecting integer overflow. The standard is published as a freely accessible online resource here: The CERT Oracle Secure Coding Standard for Java

The standard's section that describes and contains practical examples of coding techniques for preventing or detecting integer overflow is here: NUM00-J. Detect or prevent integer overflow

Book form and PDF form of The CERT Oracle Secure Coding Standard for Java are also available.

## How does Java handle integer underflows and overflows and how would yo...

Having just kinda run into this problem myself, here's my solution (for both multiplication and addition):

static boolean wouldOverflowOccurwhenMultiplying(int a, int b) { // If either a or b are Integer.MIN_VALUE, then multiplying by anything other than 0 or 1 will result in overflow if (a == 0 || b == 0) { return false; } else if (a > 0 && b > 0) { // both positive, non zero return a > Integer.MAX_VALUE / b; } else if (b < 0 && a < 0) { // both negative, non zero return a < Integer.MAX_VALUE / b; } else { // exactly one of a,b is negative and one is positive, neither are zero if (b > 0) { // this last if statements protects against Integer.MIN_VALUE / -1, which in itself causes overflow. return a < Integer.MIN_VALUE / b; } else { // a > 0 return b < Integer.MIN_VALUE / a; } } } boolean wouldOverflowOccurWhenAdding(int a, int b) { if (a > 0 && b > 0) { return a > Integer.MAX_VALUE - b; } else if (a < 0 && b < 0) { return a < Integer.MIN_VALUE - b; } return false; }

feel free to correct if wrong or if can be simplified. I've done some testing with the multiplication method, mostly edge cases, but it could still be wrong.

nice, simple solution lol

Division is apt to be slow relative to multiplication. For int*int, I'd think simply casting to long and seeing if the result fits in int would be the fastest approach. For long*long, if one normalizes operands to be positive, one can split each into upper and lower 32-bit halves, promote each half to long (be careful about sign extensions!), and then compute two partial products [one of the upper halves should be zero].

When you say "For long*long, if one normalizes operands to be positive...", how would you go about normalizing Long.MIN_VALUE?

## How does Java handle integer underflows and overflows and how would yo...

Java doesn't do anything with integer overflow for either int or long primitive types and ignores overflow with positive and negative integers.

This answer first describes the of integer overflow, gives an example of how it can happen, even with intermediate values in expression evaluation, and then gives links to resources that give detailed techniques for preventing and detecting integer overflow.

Integer arithmetic and expressions reslulting in unexpected or undetected overflow are a common programming error. Unexpected or undetected integer overflow is also a well-known exploitable security issue, especially as it affects array, stack and list objects.

Overflow can occur in either a positive or negative direction where the positive or negative value would be beyond the maximum or minimum values for the primitive type in question. Overflow can occur in an intermediate value during expression or operation evaluation and affect the outcome of an expression or operation where the final value would be expected to be within range.

Sometimes negative overflow is mistakenly called underflow. Underflow is what happens when a value would be closer to zero than the representation allows. Underflow occurs in integer arithmetic and is expected. Integer underflow happens when an integer evaluation would be between -1 and 0 or 0 and 1. What would be a fractional result truncates to 0. This is normal and expected with integer arithmetic and not considered an error. However, it can lead to code throwing an exception. One example is an "ArithmeticException: / by zero" exception if the result of integer underflow is used as a divisor in an expression.

Consider the following code:

which results in x being assigned 0 and the subsequent evaluation of bigValue / x throws an exception, "ArithmeticException: / by zero" (i.e. divide by zero), instead of y being assigned the value 2.

The expected result for x would be 858,993,458 which is less than the maximum int value of 2,147,483,647. However, the intermediate result from evaluating Integer.MAX_Value * 2, would be 4,294,967,294, which exceeds the maximum int value and is -2 in accordance with 2s complement integer representations. The subsequent evaluation of -2 / 5 evaluates to 0 which gets assigned to x.

Rearranging the expression for computing x to an expression that, when evaluated, divides before multiplying, the following code:

int bigValue = Integer.MAX_VALUE; int x = bigValue / 5 * 2; int y = bigValue / x;

results in x being assigned 858,993,458 and y being assigned 2, which is expected.

The intermediate result from bigValue / 5 is 429,496,729 which does not exceed the maximum value for an int. Subsequent evaluation of 429,496,729 * 2 doesn't exceed the maximum value for an int and the expected result gets assigned to x. The evaluation for y then does not divide by zero. The evaluations for x and y work as expected.

Java integer values are stored as and behave in accordance with 2s complement signed integer representations. When a resulting value would be larger or smaller than the maximum or minimum integer values, a 2's complement integer value results instead. In situations not expressly designed to use 2s complement behavior, which is most ordinary integer arithmetic situations, the resulting 2s complement value will cause a programming logic or computation error as was shown in the example above. An excellent Wikipedia article describes 2s compliment binary integers here: Two's complement - Wikipedia

There are techniques for avoiding unintentional integer overflow. Techinques may be categorized as using pre-condition testing, upcasting and BigInteger.

Pre-condition testing comprises examining the values going into an arithmetic operation or expression to ensure that an overflow won't occur with those values. Programming and design will need to create testing that ensures input values won't cause overflow and then determine what to do if input values occur that will cause overflow.

Upcasting comprises using a larger primitive type to perform the arithmetic operation or expression and then determining if the resulting value is beyond the maximum or minimum values for an integer. Even with upcasting, it is still possible that the value or some intermediate value in an operation or expression will be beyond the maximum or minimum values for the upcast type and cause overflow, which will also not be detected and will cause unexpected and undesired results. Through analysis or pre-conditions, it may be possible to prevent overflow with upcasting when prevention without upcasting is not possible or practical. If the integers in question are already long primitive types, then upcasting is not possible with primitive types in Java.

The BigInteger technique comprises using BigInteger for the arithmetic operation or expression using library methods that use BigInteger. BigInteger does not overflow. It will use all available memory, if necessary. Its arithmetic methods are normally only slightly less efficient than integer operations. It is still possible that a result using BigInteger may be beyond the maximum or minimum values for an integer, however, overflow will not occur in the arithmetic leading to the result. Programming and design will still need to determine what to do if a BigInteger result is beyond the maximum or minimum values for the desired primitive result type, e.g., int or long.

The Carnegie Mellon Software Engineering Institute's CERT program and Oracle have created a set of standards for secure Java programming. Included in the standards are techniques for preventing and detecting integer overflow. The standard is published as a freely accessible online resource here: The CERT Oracle Secure Coding Standard for Java

The standard's section that describes and contains practical examples of coding techniques for preventing or detecting integer overflow is here: NUM00-J. Detect or prevent integer overflow

Book form and PDF form of The CERT Oracle Secure Coding Standard for Java are also available.

## How does Java handle integer underflows and overflows and how would yo...

There are libraries that provide safe arithmetic operations, which check integer overflow/underflow . For example, Guava's IntMath.checkedAdd(int a, int b) returns the sum of a and b, provided it does not overflow, and throws ArithmeticException if a + b overflows in signed int arithmetic.

## How does Java handle integer underflows and overflows and how would yo...

Having just kinda run into this problem myself, here's my solution (for both multiplication and addition):

static boolean wouldOverflowOccurwhenMultiplying(int a, int b) { // If either a or b are Integer.MIN_VALUE, then multiplying by anything other than 0 or 1 will result in overflow if (a == 0 || b == 0) { return false; } else if (a > 0 && b > 0) { // both positive, non zero return a > Integer.MAX_VALUE / b; } else if (b < 0 && a < 0) { // both negative, non zero return a < Integer.MAX_VALUE / b; } else { // exactly one of a,b is negative and one is positive, neither are zero if (b > 0) { // this last if statements protects against Integer.MIN_VALUE / -1, which in itself causes overflow. return a < Integer.MIN_VALUE / b; } else { // a > 0 return b < Integer.MIN_VALUE / a; } } } boolean wouldOverflowOccurWhenAdding(int a, int b) { if (a > 0 && b > 0) { return a > Integer.MAX_VALUE - b; } else if (a < 0 && b < 0) { return a < Integer.MIN_VALUE - b; } return false; }

feel free to correct if wrong or if can be simplified. I've done some testing with the multiplication method, mostly edge cases, but it could still be wrong.

nice, simple solution lol

Division is apt to be slow relative to multiplication. For int*int, I'd think simply casting to long and seeing if the result fits in int would be the fastest approach. For long*long, if one normalizes operands to be positive, one can split each into upper and lower 32-bit halves, promote each half to long (be careful about sign extensions!), and then compute two partial products [one of the upper halves should be zero].

When you say "For long*long, if one normalizes operands to be positive...", how would you go about normalizing Long.MIN_VALUE?

## How does Java handle integer underflows and overflows and how would yo...

static final int safeAdd(int left, int right) throws ArithmeticException { if (right > 0 ? left > Integer.MAX_VALUE - right : left < Integer.MIN_VALUE - right) { throw new ArithmeticException("Integer overflow"); } return left + right; } static final int safeSubtract(int left, int right) throws ArithmeticException { if (right > 0 ? left < Integer.MIN_VALUE + right : left > Integer.MAX_VALUE + right) { throw new ArithmeticException("Integer overflow"); } return left - right; } static final int safeMultiply(int left, int right) throws ArithmeticException { if (right > 0 ? left > Integer.MAX_VALUE/right || left < Integer.MIN_VALUE/right : (right < -1 ? left > Integer.MIN_VALUE/right || left < Integer.MAX_VALUE/right : right == -1 && left == Integer.MIN_VALUE) ) { throw new ArithmeticException("Integer overflow"); } return left * right; } static final int safeDivide(int left, int right) throws ArithmeticException { if ((left == Integer.MIN_VALUE) && (right == -1)) { throw new ArithmeticException("Integer overflow"); } return left / right; } static final int safeNegate(int a) throws ArithmeticException { if (a == Integer.MIN_VALUE) { throw new ArithmeticException("Integer overflow"); } return -a; } static final int safeAbs(int a) throws ArithmeticException { if (a == Integer.MIN_VALUE) { throw new ArithmeticException("Integer overflow"); } return Math.abs(a); }

This handles testing. Although does not explain how Java handles integer underflows and overflows (add some text to explain).

## How does Java handle integer underflows and overflows and how would yo...

You could imagine that when you have only 2 places you are counting (so adding 1 each time)

00 01 10 11 100

But the last one gets cut down to "00" again. So there is your "overflow". You're back at 00. Now depending on what the bits mean, this can mean several things, but most of the time this means you are going from the highest value to the lowest. (11 to 00)

Mark peters adds a good one in the comments: even without overflow you'll have a problem, because the first bit is used as signing, so you'll go from high to low without losing that bit. You could say that the bit is 'separate' from the others

Well Java uses signed two's complement for most integer types so typically you are going from -1 to 0. Going from the highest value to the lowest happens the same way, but there is no bit overflow. It's when you go from 0111 to 1000, e.g.

Im not sure that exactly explains what is happening. It does explain why the operation "wraps around" but underflow/overflow have semantic meaning in the java case. In the processor the case is also "detectable" as well, and can be handled in different ways depending on the needs of the operation (set error bit, etc).