Related
This code:
System.out.println(Math.abs(Integer.MIN_VALUE));
Returns -2147483648
Should it not return the absolute value as 2147483648 ?
Integer.MIN_VALUE is -2147483648, but the highest value a 32 bit integer can contain is +2147483647. Attempting to represent +2147483648 in a 32 bit int will effectively "roll over" to -2147483648. This is because, when using signed integers, the two's complement binary representations of +2147483648 and -2147483648 are identical. This is not a problem, however, as +2147483648 is considered out of range.
For a little more reading on this matter, you might want to check out the Wikipedia article on Two's complement.
The behaviour you point out is indeed, counter-intuitive. However, this behaviour is the one specified by the javadoc for Math.abs(int):
If the argument is not negative, the argument is returned.
If the argument is negative, the negation of the argument is returned.
That is, Math.abs(int) should behave like the following Java code:
public static int abs(int x){
if (x >= 0) {
return x;
}
return -x;
}
That is, in the negative case, -x.
According to the JLS section 15.15.4, the -x is equal to (~x)+1, where ~ is the bitwise complement operator.
To check whether this sounds right, let's take -1 as example.
The integer value -1 is can be noted as 0xFFFFFFFF in hexadecimal in Java (check this out with a println or any other method). Taking -(-1) thus gives:
-(-1) = (~(0xFFFFFFFF)) + 1 = 0x00000000 + 1 = 0x00000001 = 1
So, it works.
Let us try now with Integer.MIN_VALUE . Knowing that the lowest integer can be represented by 0x80000000, that is, the first bit set to 1 and the 31 remaining bits set to 0, we have:
-(Integer.MIN_VALUE) = (~(0x80000000)) + 1 = 0x7FFFFFFF + 1
= 0x80000000 = Integer.MIN_VALUE
And this is why Math.abs(Integer.MIN_VALUE) returns Integer.MIN_VALUE. Also note that 0x7FFFFFFF is Integer.MAX_VALUE.
That said, how can we avoid problems due to this counter-intuitive return value in the future?
We could, as pointed out by #Bombe, cast our ints to long before. We, however, must either
cast them back into ints, which does not work because
Integer.MIN_VALUE == (int) Math.abs((long)Integer.MIN_VALUE).
Or continue with longs somehow hoping that we'll never call Math.abs(long) with a value equal to Long.MIN_VALUE, since we also have Math.abs(Long.MIN_VALUE) == Long.MIN_VALUE.
We can use BigIntegers everywhere, because BigInteger.abs() does indeed always return a positive value. This is a good alternative, though a bit slower than manipulating raw integer types.
We can write our own wrapper for Math.abs(int), like this:
/**
* Fail-fast wrapper for {#link Math#abs(int)}
* #param x
* #return the absolute value of x
* #throws ArithmeticException when a negative value would have been returned by {#link Math#abs(int)}
*/
public static int abs(int x) throws ArithmeticException {
if (x == Integer.MIN_VALUE) {
// fail instead of returning Integer.MAX_VALUE
// to prevent the occurrence of incorrect results in later computations
throw new ArithmeticException("Math.abs(Integer.MIN_VALUE)");
}
return Math.abs(x);
}
Use a integer bitwise AND to clear the high bit, ensuring that the result is non-negative: int positive = value & Integer.MAX_VALUE (essentially overflowing from Integer.MAX_VALUE to 0 instead of Integer.MIN_VALUE)
As a final note, this problem seems to be known for some time. See for example this entry about the corresponding findbugs rule.
Here is what Java doc says for Math.abs() in javadoc:
Note that if the argument is equal to
the value of Integer.MIN_VALUE, the
most negative representable int value,
the result is that same value, which
is negative.
To see the result that you are expecting, cast Integer.MIN_VALUE to long:
System.out.println(Math.abs((long) Integer.MIN_VALUE));
There is a fix to this in Java 15 will be a method to int and long. They will be present on the classes
java.lang.Math and java.lang.StrictMath
The methods.
public static int absExact(int a)
public static long absExact(long a)
If you pass
Integer.MIN_VALUE
OR
Long.MIN_VALUE
A Exception is thrown.
https://bugs.openjdk.java.net/browse/JDK-8241805
I would like to see if either Long.MIN_VALUE or Integer.MIN_VALUE is passed a positive value would be return and not a exception but.
2147483648 cannot be stored in an integer in java, its binary representation is the same as -2147483648.
But (int) 2147483648L == -2147483648 There is one negative number which has no positive equivalent so there is not positive value for it. You will see the same behaviour with Long.MAX_VALUE.
Math.abs doesn't work all the time with big numbers I use this little code logic that I learnt when I was 7 years old!
if(Num < 0){
Num = -(Num);
}
Is there any way to find the absolute value of a number without using the Math.abs() method in java.
If you look inside Math.abs you can probably find the best answer:
Eg, for floats:
/*
* Returns the absolute value of a {#code float} value.
* If the argument is not negative, the argument is returned.
* If the argument is negative, the negation of the argument is returned.
* Special cases:
* <ul><li>If the argument is positive zero or negative zero, the
* result is positive zero.
* <li>If the argument is infinite, the result is positive infinity.
* <li>If the argument is NaN, the result is NaN.</ul>
* In other words, the result is the same as the value of the expression:
* <p>{#code Float.intBitsToFloat(0x7fffffff & Float.floatToIntBits(a))}
*
* #param a the argument whose absolute value is to be determined
* #return the absolute value of the argument.
*/
public static float abs(float a) {
return (a <= 0.0F) ? 0.0F - a : a;
}
Yes:
abs_number = (number < 0) ? -number : number;
For integers, this works fine (except for Integer.MIN_VALUE, whose absolute value cannot be represented as an int).
For floating-point numbers, things are more subtle. For example, this method -- and all other methods posted thus far -- won't handle the negative zero correctly.
To avoid having to deal with such subtleties yourself, my advice would be to stick to Math.abs().
Like this:
if (number < 0) {
number *= -1;
}
Since Java is a statically typed language, I would expect that a abs-method which takes an int returns an int, if it expects a float returns a float, for a Double, return a Double. Maybe it could return always the boxed or unboxed type for doubles and Doubles and so on.
So you need one method per type, but now you have a new problem: For byte, short, int, long the range for negative values is 1 bigger than for positive values.
So what should be returned for the method
byte abs (byte in) {
// #todo
}
If the user calls abs on -128? You could always return the next bigger type so that the range is guaranteed to fit to all possible input values. This will lead to problems for long, where no normal bigger type exists, and make the user always cast the value down after testing - maybe a hassle.
The second option is to throw an arithmetic exception. This will prevent casting and checking the return type for situations where the input is known to be limited, such that X.MIN_VALUE can't happen. Think of MONTH, represented as int.
byte abs (byte in) throws ArithmeticException {
if (in == Byte.MIN_VALUE) throw new ArithmeticException ("abs called on Byte.MIN_VALUE");
return (in < 0) ? (byte) -in : in;
}
The "let's ignore the rare cases of MIN_VALUE" habit is not an option. First make the code work - then make it fast. If the user needs a faster, but buggy solution, he should write it himself.
The simplest solution that might work means: simple, but not too simple.
Since the code doesn't rely on state, the method can and should be made static. This allows for a quick test:
public static void main (String args []) {
System.out.println (abs(new Byte ( "7")));
System.out.println (abs(new Byte ("-7")));
System.out.println (abs((byte) 7));
System.out.println (abs((byte) -7));
System.out.println (abs(new Byte ( "127")));
try
{
System.out.println (abs(new Byte ("-128")));
}
catch (ArithmeticException ae)
{
System.out.println ("Integer: " + Math.abs (new Integer ("-128")));
}
System.out.println (abs((byte) 127));
System.out.println (abs((byte) -128));
}
I catch the first exception and let it run into the second, just for demonstration.
There is a bad habit in programming, which is that programmers care much more for fast than for correct code. What a pity!
If you're curious why there is one more negative than positive value, I have a diagram for you.
Although this shouldn't be a bottle neck as branching issues on modern processors isn't normally a problem, but in the case of integers you could go for a branch-less solution as outlined here: http://graphics.stanford.edu/~seander/bithacks.html#IntegerAbs.
(x + (x >> 31)) ^ (x >> 31);
This does fail in the obvious case of Integer.MIN_VALUE however, so this is a use at your own risk solution.
In case of the absolute value of an integer x without using Math.abs(), conditions or bit-wise operations, below could be a possible solution in Java.
(int)(((long)x*x - 1)%(double)x + 1);
Because Java treats a%b as a - a/b * b, the sign of the result will be same as "a" no matter what sign of "b" is; (x*x-1)%x will equal abs(x)-1; type casting of "long" is to prevent overflow and double allows dividing by zero.
Again, x = Integer.MIN_VALUE will cause overflow due to subtracting 1.
You can use :
abs_num = (num < 0) ? -num : num;
Here is a one-line solution that will return the absolute value of a number:
abs_number = (num < 0) ? -num : num;
-num will equal to num for Integer.MIN_VALUE as
Integer.MIN_VALUE = Integer.MIN_VALUE * -1
Lets say if N is the number for which you want to calculate the absolute value(+ve number( without sign))
if (N < 0)
{
N = (-1) * N;
}
N will now return the Absolute value
Is there a Java function to convert a positive int to a negative one and a negative int to a positive one?
I'm looking for a reverse function to perform this conversion:
-5 -> 5
5 -> -5
What about x *= -1; ? Do you really want a library function for this?
x = -x;
This is probably the most trivial question I have ever seen anywhere.
... and why you would call this trivial function 'reverse()' is another mystery.
Just use the unary minus operator:
int x = 5;
...
x = -x; // Here's the mystery library function - the single character "-"
Java has two minus operators:
the familiar arithmetic version (eg 0 - x), and
the unary minus operation (used here), which negates the (single) operand
This compiles and works as expected.
Another method (2's complement):
public int reverse(int x){
x~=x;
x++;
return x;
}
It does a one's complement first (by complementing all the bits) and then adds 1 to x. This method does the job as well.
Note: This method is written in Java, and will be similar to a lot of other languages
No such function exists or is possible to write.
The problem is the edge case Integer.MIN_VALUE (-2,147,483,648 = 0x80000000) apply each of the three methods above and you get the same value out. This is due to the representation of integers and the maximum possible integer Integer.MAX_VALUE (-2,147,483,647 = 0x7fffffff) which is one less what -Integer.MIN_VALUE should be.
Yes, as was already noted by Jeffrey Bosboom (Sorry Jeffrey, I hadn't noticed your comment when I answered), there is such a function: Math.negateExact.
and
No, you probably shouldn't be using it. Not unless you need a method reference.
original *= -1;
Simple line of code, original is any int you want it to be.
Necromancing here.
Obviously, x *= -1; is far too simple.
Instead, we could use a trivial binary complement:
number = ~(number - 1) ;
Like this:
import java.io.*;
/* Name of the class has to be "Main" only if the class is public. */
class Ideone
{
public static void main (String[] args) throws java.lang.Exception
{
int iPositive = 15;
int iNegative = ( ~(iPositive - 1) ) ; // Use extra brackets when using as C preprocessor directive ! ! !...
System.out.println(iNegative);
iPositive = ~(iNegative - 1) ;
System.out.println(iPositive);
iNegative = 0;
iPositive = ~(iNegative - 1);
System.out.println(iPositive);
}
}
That way we can ensure that mediocre programmers don't understand what's going on ;)
The easiest thing to do is 0- the value
for instance if int i = 5;
0-i would give you -5
and if i was -6;
0- i would give you 6
You can use the minus operator or Math.abs. These work for all negative integers EXCEPT for Integer.MIN_VALUE!
If you do 0 - MIN_VALUE the answer is still MIN_VALUE.
For converting a negative number to positive. Simply use Math.abs() inbuilt function.
int n = -10;
n = Math.abs(n);
All the best!
In kotlin you can use unaryPlus and unaryMinus
input = input.unaryPlus()
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-minus.html
You can use Math:
int x = Math.abs(-5);
I have two Numbers. Eg:
Number a = 2;
Number b = 3;
//Following is an error:
Number c = a + b;
Why arithmetic operations are not supported on Numbers? Anyway how would I add these two numbers in java? (Of course I'm getting them from somewhere and I don't know if they are Integer or float etc).
You say you don't know if your numbers are integer or float... when you use the Number class, the compiler also doesn't know if your numbers are integers, floats or some other thing. As a result, the basic math operators like + and - don't work; the computer wouldn't know how to handle the values.
START EDIT
Based on the discussion, I thought an example might help. Computers store floating point numbers as two parts, a coefficient and an exponent. So, in a theoretical system, 001110 might be broken up as 0011 10, or 32 = 9. But positive integers store numbers as binary, so 001110 could also mean 2 + 4 + 8 = 14. When you use the class Number, you're telling the computer you don't know if the number is a float or an int or what, so it knows it has 001110 but it doesn't know if that means 9 or 14 or some other value.
END EDIT
What you can do is make a little assumption and convert to one of the types to do the math. So you could have
Number c = a.intValue() + b.intValue();
which you might as well turn into
Integer c = a.intValue() + b.intValue();
if you're willing to suffer some rounding error, or
Float c = a.floatValue() + b.floatValue();
if you suspect that you're not dealing with integers and are okay with possible minor precision issues. Or, if you'd rather take a small performance blow instead of that error,
BigDecimal c = new BigDecimal(a.floatValue()).add(new BigDecimal(b.floatValue()));
It would also work to make a method to handle the adding for you. Now I do not know the performance impact this will cause but I assume it will be less than using BigDecimal.
public static Number addNumbers(Number a, Number b) {
if(a instanceof Double || b instanceof Double) {
return a.doubleValue() + b.doubleValue();
} else if(a instanceof Float || b instanceof Float) {
return a.floatValue() + b.floatValue();
} else if(a instanceof Long || b instanceof Long) {
return a.longValue() + b.longValue();
} else {
return a.intValue() + b.intValue();
}
}
The only way to correctly add any two types of java.lang.Number is:
Number a = 2f; // Foat
Number b = 3d; // Double
Number c = new BigDecimal( a.toString() ).add( new BigDecimal( b.toString() ) );
This works even for two arguments with a different number-type. It will (should?) not produce any sideeffects like overflows or loosing precision, as far as the toString() of the number-type does not reduce precision.
java.lang.Number is just the superclass of all wrapper classes of primitive types (see java doc). Use the appropriate primitive type (double, int, etc.) for your purpose, or the respective wrapper class (Double, Integer, etc.).
Consider this:
Number a = 1.5; // Actually Java creates a double and boxes it into a Double object
Number b = 1; // Same here for int -> Integer boxed
// What should the result be? If Number would do implicit casts,
// it would behave different from what Java usually does.
Number c = a + b;
// Now that works, and you know at first glance what that code does.
// Nice explicit casts like you usually use in Java.
// The result is of course again a double that is boxed into a Double object
Number d = a.doubleValue() + (double)b.intValue();
Use the following:
Number c = a.intValue() + b.intValue(); // Number is an object and not a primitive data type.
Or:
int a = 2;
int b = 3;
int c = 2 + 3;
I think there are 2 sides to your question.
Why is operator+ not supported on Number?
Because the Java language spec. does not specify this, and there is no operator overloading. There is also not a compile-time natural way to cast the Number to some fundamental type, and there is no natural add to define for some type of operations.
Why are basic arithmic operations not supported on Number?
(Copied from my comment:)
Not all subclasses can implement this in a way you would expect. Especially with the Atomic types it's hard to define a usefull contract for e.g. add.
Also, a method add would be trouble if you try to add a Long to a Short.
If you know the Type of one number but not the other it is possible to do something like
public Double add(Double value, Number increment) {
return value + Double.parseDouble(increment.toString());
}
But it can be messy, so be aware of potential loss of accuracy and NumberFormatExceptions
Number is an abstract class which you cannot make an instance of. Provided you have a correct instance of it, you can get number.longValue() or number.intValue() and add them.
First of all, you should be aware that Number is an abstract class. What happens here is that when you create your 2 and 3, they are interpreted as primitives and a subtype is created (I think an Integer) in that case. Because an Integer is a subtype of Number, you can assign the newly created Integer into a Number reference.
However, a number is just an abstraction. It could be integer, it could be floating point, etc., so the semantics of math operations would be ambiguous.
Number does not provide the classic map operations for two reasons:
First, member methods in Java cannot be operators. It's not C++. At best, they could provide an add()
Second, figuring out what type of operation to do when you have two inputs (e.g., a division of a float by an int) is quite tricky.
So instead, it is your responsibility to make the conversion back to the specific primitive type you are interested in it and apply the mathematical operators.
The best answer would be to make util with double dispatch drilling down to most known types (take a look at Smalltalk addtition implementation)
I have a Java method in which I'm summing a set of numbers. However, I want any negatives numbers to be treated as positives. So (1)+(2)+(1)+(-1) should equal 5.
I'm sure there is very easy way of doing this - I just don't know how.
Just call Math.abs. For example:
int x = Math.abs(-5);
Which will set x to 5.
Note that if you pass Integer.MIN_VALUE, the same value (still negative) will be returned, as the range of int does not allow the positive equivalent to be represented.
The concept you are describing is called "absolute value", and Java has a function called Math.abs to do it for you. Or you could avoid the function call and do it yourself:
number = (number < 0 ? -number : number);
or
if (number < 0)
number = -number;
You're looking for absolute value, mate. Math.abs(-5) returns 5...
Use the abs function:
int sum=0;
for(Integer i : container)
sum+=Math.abs(i);
Try this (the negative in front of the x is valid since it is a unary operator, find more here):
int answer = -x;
With this, you can turn a positive to a negative and a negative to a positive.
However, if you want to only make a negative number positive then try this:
int answer = Math.abs(x);
A little cool math trick! Squaring the number will guarantee a positive value of x^2, and then, taking the square root will get you to the absolute value of x:
int answer = Math.sqrt(Math.pow(x, 2));
Hope it helps! Good Luck!
This code is not safe to be called on positive numbers.
int x = -20
int y = x + (2*(-1*x));
// Therefore y = -20 + (40) = 20
Are you asking about absolute values?
Math.abs(...) is the function you probably want.
You want to wrap each number into Math.abs(). e.g.
System.out.println(Math.abs(-1));
prints out "1".
If you want to avoid writing the Math.-part, you can include the Math util statically. Just write
import static java.lang.Math.abs;
along with your imports, and you can refer to the abs()-function just by writing
System.out.println(abs(-1));
The easiest, if verbose way to do this is to wrap each number in a Math.abs() call, so you would add:
Math.abs(1) + Math.abs(2) + Math.abs(1) + Math.abs(-1)
with logic changes to reflect how your code is structured. Verbose, perhaps, but it does what you want.
When you need to represent a value without the concept of a loss or absence (negative value), that is called "absolute value".
The logic to obtain the absolute value is very simple: "If it's positive, maintain it. If it's negative, negate it".
What this means is that your logic and code should work like the following:
//If value is negative...
if ( value < 0 ) {
//...negate it (make it a negative negative-value, thus a positive value).
value = negate(value);
}
There are 2 ways you can negate a value:
By, well, negating it's value: value = (-value);
By multiplying it by "100% negative", or "-1": value = value *
(-1);
Both are actually two sides of the same coin. It's just that you usually don't remember that value = (-value); is actually value = 1 * (-value);.
Well, as for how you actually do it in Java, it's very simple, because Java already provides a function for that, in the Math class: value = Math.abs(value);
Yes, doing it without Math.abs() is just a line of code with very simple math, but why make your code look ugly? Just use Java's provided Math.abs() function! They provide it for a reason!
If you absolutely need to skip the function, you can use value = (value < 0) ? (-value) : value;, which is simply a more compact version of the code I mentioned in the logic (3rd) section, using the Ternary operator (? :).
Additionally, there might be situations where you want to always represent loss or absence within a function that might receive both positive and negative values.
Instead of doing some complicated check, you can simply get the absolute value, and negate it: negativeValue = (-Math.abs(value));
With that in mind, and considering a case with a sum of multiple numbers such as yours, it would be a nice idea to implement a function:
int getSumOfAllAbsolutes(int[] values){
int total = 0;
for(int i=0; i<values.lenght; i++){
total += Math.abs(values[i]);
}
return total;
}
Depending on the probability you might need related code again, it might also be a good idea to add them to your own "utils" library, splitting such functions into their core components first, and maintaining the final function simply as a nest of calls to the core components' now-split functions:
int[] makeAllAbsolute(int[] values){
//#TIP: You can also make a reference-based version of this function, so that allocating 'absolutes[]' is not needed, thus optimizing.
int[] absolutes = values.clone();
for(int i=0; i<values.lenght; i++){
absolutes[i] = Math.abs(values[i]);
}
return absolutes;
}
int getSumOfAllValues(int[] values){
int total = 0;
for(int i=0; i<values.lenght; i++){
total += values[i];
}
return total;
}
int getSumOfAllAbsolutes(int[] values){
return getSumOfAllValues(makeAllAbsolute(values));
}
Why don't you multiply that number with -1?
Like This:
//Given x as the number, if x is less than 0, return 0 - x, otherwise return x:
return (x <= 0.0F) ? 0.0F - x : x;
If you're interested in the mechanics of two's complement, here's the absolutely inefficient, but illustrative low-level way this is made:
private static int makeAbsolute(int number){
if(number >=0){
return number;
} else{
return (~number)+1;
}
}
Library function Math.abs() can be used.
Math.abs() returns the absolute value of the argument
if the argument is negative, it returns the negation of the argument.
if the argument is positive, it returns the number as it is.
e.g:
int x=-5;
System.out.println(Math.abs(x));
Output: 5
int y=6;
System.out.println(Math.abs(y));
Output: 6
String s = "-1139627840";
BigInteger bg1 = new BigInteger(s);
System.out.println(bg1.abs());
Alternatively:
int i = -123;
System.out.println(Math.abs(i));
To convert negative number to positive number (this is called absolute value), uses Math.abs(). This Math.abs() method is work like this
“number = (number < 0 ? -number : number);".
In below example, Math.abs(-1) will convert the negative number 1 to positive 1.
example
public static void main(String[] args) {
int total = 1 + 1 + 1 + 1 + (-1);
//output 3
System.out.println("Total : " + total);
int total2 = 1 + 1 + 1 + 1 + Math.abs(-1);
//output 5
System.out.println("Total 2 (absolute value) : " + total2);
}
Output
Total : 3
Total 2 (absolute value) : 5
I would recommend the following solutions:
without lib fun:
value = (value*value)/value
(The above does not actually work.)
with lib fun:
value = Math.abs(value);
I needed the absolute value of a long , and looked deeply into Math.abs and found that if my argument is less than LONG.MIN_VAL which is -9223372036854775808l, then the abs function would not return an absolute value but only the minimum value. Inthis case if your code is using this abs value further then there might be an issue.
Can you please try this one?
public static int toPositive(int number) {
return number & 0x7fffffff;
}
if(arr[i]<0)
Math.abs(arr[i]); //1st way (taking absolute value)
arr[i]=-(arr[i]); //2nd way (taking -ve of -ve no. yields a +ve no.)
arr[i]= ~(arr[i]-1); //3rd way (taking negation)
I see people are saying that Math.abs(number) but this method is not full proof.
This fails when you try to wrap Math.abs(Integer.MIN_VALUE) (see ref. https://youtu.be/IWrpDP-ad7g)
If you are not sure whether you are going to receive the Integer.MIN_VALUE in the input. It is always recommended to check for that number and handle it manually.
In kotlin you can use unaryPlus
input = input.unaryPlus()
https://kotlinlang.org/api/latest/jvm/stdlib/kotlin/-int/unary-plus.html
Try this in the for loop:
sum += Math.abs(arr[i])
dont do this
number = (number < 0 ? -number : number);
or
if (number < 0) number = -number;
this will be an bug when you run find bug on your code it will report it as RV_NEGATING_RESULT_OF