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I'm making a small program that determines the minimum integer of 3 integers. I'm having problems returning the integers back into the variable answer.
Here Is how I imagine the program working;
PROGRAM RUNS:
Looks for Method called "Minimumum" with the listed argument.
Determines the minimum integer and returns it back to the method Minimum
This value gets stored in the answer variable
Code:
public class Method {
public static void main(String[] args) {
int answer = Minimum(20, 40, 50);
}
public static int Minimum(int first, int second, int third) {
if ((first < (second) && (first < third))) {
return first;
} else if ((second < (first) && (second < third))) {
return second;
} else if (((third < first) && (third < second))) {
return (third);
} else {
System.out.println("error");
}
}
}
You need to return a result in all cases, so your else block is incorrect. Do you really think there is a fourth case ? No, there isn't : there are three integers so the minimum is one of those three, meaning there only are 3 cases... Simply remove the else block. By the way, why do you use so many parentheses ? It's useless, and unreadable.
public static int Minimum(int first, int second, int third){
if (first < second && first < third)
return first;
else if (second < first && second < third)
return second;
else
return third;
}
As noted by the others, this is not sufficient to make your method correct. Indeed, you don't care about strict inequality here because when two numbers are the same, you can choose either of them as the minimum. This code breaks if the two first parameters are equal. To fix it, simply use <= instead of < (everywhere).
Your code looks a bit complicated.
int Minimum(int first, int second, int third) {
int min = first;
if (second < min) {
min = second;
}
if (third < min) {
min = third;
}
return min;
}
This looks creepy, but it should do it.
Hope it helps to understand.
You have to return in your final else block. But you could simplify your code with Math.min(int, int). Something like,
public static int Minimum(int first, int second, int third) {
return Math.min(first, Math.min(second, third));
}
Minimum function should always return the integer value in any case. After else, add return statement.
In Java 8+ your method could also look as follows:
public static int minimum(Integer val1, Integer val2, Integer val3){
List<Integer> list = Arrays.asList(new Integer[]{val1, val2, val3});
return list.stream().min((Integer v1, Integer v2) -> v1 < v2 ? 0 : 1 ).get();
}
I have written multiple attempts to this problem, but I think this is the closest I could get. This solution makes the method recurse infinitely, because I don't have a base case, and I can't figure it out. The counter++ line is unreachable, and I can't get this to work, and I am very tired. This would be very easy with a loop, but recursion is kind of a new concept to me, and I would be thankful if someone helped me solve this.
public static double pi(int a, double b){
int counter=0;
if (counter %2==0){
return a-(a/(pi(a,b+2)));
counter++;
} else {
return a+(a/(pi(a,b+2)));
counter++;
}
You could pass in another int, say limit, and add this code:
if (b > limit) {
return a;
}
Or you could pass in some tolerance value:
if (pi(a,b+2) < tolerance) {
return a;
}
Whenever you're working with recursion it's good to establish an exit strategy up front.
Here is an implementation that works. Do not use it:
public static double term(double acc, int n, int r) {
if (r-- > 0) {
double sgn = (n % 4 == 1) ? +1.0 : -1.0;
acc += sgn * 4.0 / n;
n += 2;
return term(acc, n, r);
} else {
return acc;
}
}
public static double pi() {
return term(0.0, 1, 1000);
}
The reason not to use it is that this particular infinite series is a particularly poor way of calculating π because it converges very slowly. In the example above event after 1000 iterations are performed it's still only correct to 3 decimal places because the final calculated term is 4 / 1000.
Going much beyond 1000 iterations results in a stack overflow error with little improvement in the accuracy even though the term function is (I think) potentially tail recursive.
Is it possible to do this?
double variable;
variable = 5;
/* the below should return true, since 5 is an int.
if variable were to equal 5.7, then it would return false. */
if(variable == int) {
//do stuff
}
I know the code probably doesn't go anything like that, but how does it go?
Or you could use the modulo operator:
(d % 1) == 0
if ((variable == Math.floor(variable)) && !Double.isInfinite(variable)) {
// integer type
}
This checks if the rounded-down value of the double is the same as the double.
Your variable could have an int or double value and Math.floor(variable) always has an int value, so if your variable is equal to Math.floor(variable) then it must have an int value.
This also doesn't work if the value of the variable is infinite or negative infinite hence adding 'as long as the variable isn't inifinite' to the condition.
Guava: DoubleMath.isMathematicalInteger. (Disclosure: I wrote it.) Or, if you aren't already importing Guava, x == Math.rint(x) is the fastest way to do it; rint is measurably faster than floor or ceil.
public static boolean isInt(double d)
{
return d == (int) d;
}
Try this way,
public static boolean isInteger(double number){
return Math.ceil(number) == Math.floor(number);
}
for example:
Math.ceil(12.9) = 13; Math.floor(12.9) = 12;
hence 12.9 is not integer, nevertheless
Math.ceil(12.0) = 12; Math.floor(12.0) =12;
hence 12.0 is integer
Here is a good solution:
if (variable == (int)variable) {
//logic
}
Consider:
Double.isFinite (value) && Double.compare (value, StrictMath.rint (value)) == 0
This sticks to core Java and avoids an equality comparison between floating point values (==) which is consdered bad. The isFinite() is necessary as rint() will pass-through infinity values.
Here's a version for Integer and Double:
private static boolean isInteger(Double variable) {
if ( variable.equals(Math.floor(variable)) &&
!Double.isInfinite(variable) &&
!Double.isNaN(variable) &&
variable <= Integer.MAX_VALUE &&
variable >= Integer.MIN_VALUE) {
return true;
} else {
return false;
}
}
To convert Double to Integer:
Integer intVariable = variable.intValue();
Similar to SkonJeet's answer above, but the performance is better (at least in java):
Double zero = 0d;
zero.longValue() == zero.doubleValue()
My simple solution:
private boolean checkIfInt(double value){
return value - Math.floor(value) == 0;
}
public static boolean isInteger(double d) {
// Note that Double.NaN is not equal to anything, even itself.
return (d == Math.floor(d)) && !Double.isInfinite(d);
}
A simple way for doing this could be
double d = 7.88; //sample example
int x=floor(d); //floor of number
int y=ceil(d); //ceil of number
if(x==y) //both floor and ceil will be same for integer number
cout<<"integer number";
else
cout<<"double number";
My solution would be
double variable=the number;
if(variable-(int)variable=0.0){
// do stuff
}
you could try in this way: get the integer value of the double, subtract this from the original double value, define a rounding range and tests if the absolute number of the new double value(without the integer part) is larger or smaller than your defined range. if it is smaller you can intend it it is an integer value. Example:
public final double testRange = 0.2;
public static boolean doubleIsInteger(double d){
int i = (int)d;
double abs = Math.abs(d-i);
return abs <= testRange;
}
If you assign to d the value 33.15 the method return true. To have better results you can assign lower values to testRange (as 0.0002) at your discretion.
Personally, I prefer the simple modulo operation solution in the accepted answer.
Unfortunately, SonarQube doesn't like equality tests with floating points without setting a round precision. So we have tried to find a more compliant solution. Here it is:
if (new BigDecimal(decimalValue).remainder(new BigDecimal(1)).equals(BigDecimal.ZERO)) {
// no decimal places
} else {
// decimal places
}
Remainder(BigDecimal) returns a BigDecimal whose value is (this % divisor). If this one's equal to zero, we know there is no floating point.
Because of % operator cannot apply to BigDecimal and int (i.e. 1) directly, so I am using the following snippet to check if the BigDecimal is an integer:
value.stripTrailingZeros().scale() <= 0
Similar (and probably inferior) to Eric Tan's answer (which checks scale):
double d = 4096.00000;
BigDecimal bd = BigDecimal.valueOf(d);
String s = bd.stripTrailingZeros().toPlainString();
boolean isInteger = s.indexOf(".")==-1;
Here's a solution:
float var = Your_Value;
if ((var - Math.floor(var)) == 0.0f)
{
// var is an integer, so do stuff
}
I have written the following code:
public class NewClass2 implements Comparator<Point>
{
public int compare(Point p1, Point p2)
{
return (int)(p1.getY() - p2.getY());
}
}
If I let's say have two double numbers, 3.2 - 3.1, the difference should be 0.1. When I cast the number to an int, however, the difference ends up as 0, which is not correct.
I therefore need compare() to return a double, not an int. The problem is, my getX field is a double. How can I solve this problem?
I suggest you use the builtin method Double.compare(). If you need a range for double values to be equal you can use chcek for that first.
return Double.compare(p1.getY(), p2.gety());
or
if(Math.abs(p1.getY()-p2.getY()) < ERR) return 0;
return Double.compare(p1.getY(), p2.gety());
The problem with using < and > is that NaN will return false in both cases resulting in a possibly inconsistent handling. e.g. NaN is defined as not being equal to anything, even itself however in #suihock's and #Martinho's solutions, if either value is NaN the method will return 0 everytime, implying that NaN is equal to everything.
You don't need to return double.
The Comparator interface is used to establish an ordering for the elements being compared. Having fields that use double is irrelevant to this ordering.
Your code is fine.
Sorry, I was wrong, reading the question again, this is what you need:
public class NewClass2 implements Comparator<Point> {
public int compare(Point p1, Point p2) {
if (p1.getY() < p2.getY()) return -1;
if (p1.getY() > p2.getY()) return 1;
return 0;
}
}
Since Java 1.8 you can also use
Comparator.comparingDouble(p -> p.getY())
The method compare should return an int. It is a number that is either:
Less than zero, if the first value is less than the second;
Equal to zero, if the two values are equal;
Greater than zero, if the first value is greater than the second;
You don't need to return a double. You must return an int to implement the Comparator interface. You just have to return the correct int, according to the rules I outlined above.
You can't simply cast from int, as, like you said, a difference of 0.1 will result in 0. You can simply do this:
public int compare(Point p1, Point p2)
{
double delta= p1.getY() - p2.getY();
if(delta > 0) return 1;
if(delta < 0) return -1;
return 0;
}
But since comparison of floating-point values is always troublesome, you should compare within a certain range (see this question), something like this:
public int compare(Point p1, Point p2)
{
double delta = p1.getY() - p2.getY();
if(delta > 0.00001) return 1;
if(delta < -0.00001) return -1;
return 0;
}
I just want to expand on Peter Lawrey answer on JDK 8, if you do it like this:
public class NewClass2 implements Comparator<Point> {
public int compare(Point p1, Point p2) {
return Double.compare(p1.getY(), p2.gety());
}
}
You could define this comparator using a lambda expression pretty easily
(Point p1,Point p2) -> Double.compare(p1.getY(), p2.gety())
Better yet, you could use a member reference like this:
Double::compare
It is so convinent in Java 8, choose anyone just as you wish:
Comparator<someClass> cp = (a, b) -> Double.compare(a.getScore(), b.getScore());
Comparator<someClass> cp = Comparator.comparing(someClass::getScore);
Comparator<someClass> cp = Comparator.comparingDouble(someClass::getScore);
int compare(Double first, Double second) {
if (Math.abs(first - second) < 1E-6) {
return 0;
} else {
return Double.compare(first, second);
}
}
Use Double.compare(/**double value 1*/, /**double value 2*/); with a new Comparator for your model class double value.
public static List<MyModel> sortByDouble(List<MyModel> modelList) {
Collections.sort(modelList, new Comparator<MyModel>() {
#Override
public int compare(MyModels1, MyModels2) {
double s1Distance = Double.parseDouble(!TextUtils.isEmpty(s1.distance) ? s1.distance : "0");
double s2Distance = Double.parseDouble(!TextUtils.isEmpty(s2.distance) ? s2.distance : "0");
return Double.compare(s1Distance, s2Distance);
}
});
return modelList;
}
Well, you could multiply those double values by an appropriate factor before converting into integer, for eg. in your case since its only one decimal place so 10 would be a good factor;
return (int)(p1.getY()*10 - p2.getY()*10);
Double min = Arrays.stream(myArray).min(Double::compare).get();
I was asked in an interview, how to determine whether a number is positive or negative. The rules are that we should not use relational operators such as <, and >, built in java functions (like substring, indexOf, charAt, and startsWith), no regex, or API's.
I did some homework on this and the code is given below, but it only works for integer type. But they asked me to write a generic code that works for float, double, and long.
// This might not be better way!!
S.O.P ((( number >> 31 ) & 1) == 1 ? "- ve number " : "+ve number );
any ideas from your side?
The integer cases are easy. The double case is trickier, until you remember about infinities.
Note: If you consider the double constants "part of the api", you can replace them with overflowing expressions like 1E308 * 2.
int sign(int i) {
if (i == 0) return 0;
if (i >> 31 != 0) return -1;
return +1;
}
int sign(long i) {
if (i == 0) return 0;
if (i >> 63 != 0) return -1;
return +1;
}
int sign(double f) {
if (f != f) throw new IllegalArgumentException("NaN");
if (f == 0) return 0;
f *= Double.POSITIVE_INFINITY;
if (f == Double.POSITIVE_INFINITY) return +1;
if (f == Double.NEGATIVE_INFINITY) return -1;
//this should never be reached, but I've been wrong before...
throw new IllegalArgumentException("Unfathomed double");
}
The following is a terrible approach that would get you fired at any job...
It depends on you getting a Stack Overflow Exception [or whatever Java calls it]... And it would only work for positive numbers that don't deviate from 0 like crazy.
Negative numbers are fine, since you would overflow to positive, and then get a stack overflow exception eventually [which would return false, or "yes, it is negative"]
Boolean isPositive<T>(T a)
{
if(a == 0) return true;
else
{
try
{
return isPositive(a-1);
}catch(StackOverflowException e)
{
return false; //It went way down there and eventually went kaboom
}
}
}
This will only works for everything except [0..2]
boolean isPositive = (n % (n - 1)) * n == n;
You can make a better solution like this (works except for [0..1])
boolean isPositive = ((n % (n - 0.5)) * n) / 0.5 == n;
You can get better precision by changing the 0.5 part with something like 2^m (m integer):
boolean isPositive = ((n % (n - 0.03125)) * n) / 0.03125 == n;
You can do something like this:
((long) (num * 1E308 * 1E308) >> 63) == 0 ? "+ve" : "-ve"
The main idea here is that we cast to a long and check the value of the most significant bit. As a double/float between -1 and 0 will round to zero when cast to a long, we multiply by large doubles so that a negative float/double will be less than -1. Two multiplications are required because of the existence of subnormals (it doesn't really need to be that big though).
What about this?
return ((num + "").charAt(0) == '-');
// Returns 0 if positive, nonzero if negative
public long sign(long value) {
return value & 0x8000000000000000L;
}
Call like:
long val1 = ...;
double val2 = ...;
float val3 = ...;
int val4 = ...;
sign((long) valN);
Casting from double / float / integer to long should preserve the sign, if not the actual value...
You say
we should not use conditional operators
But this is a trick requirement, because == is also a conditional operator. There is also one built into ? :, while, and for loops. So nearly everyone has failed to provide an answer meeting all the requirements.
The only way to build a solution without a conditional operator is to use lookup table vs one of a few other people's solutions that can be boiled down to 0/1 or a character, before a conditional is met.
Here are the answers that I think might work vs a lookup table:
Nabb
Steven Schlansker
Dennis Cheung
Gary Rowe
This solution uses modulus. And yes, it also works for 0.5 (tests are below, in the main method).
public class Num {
public static int sign(long x) {
if (x == 0L || x == 1L) return (int) x;
return x == Long.MIN_VALUE || x % (x - 1L) == x ? -1 : 1;
}
public static int sign(double x) {
if (x != x) throw new IllegalArgumentException("NaN");
if (x == 0.d || x == 1.d) return (int) x;
if (x == Double.POSITIVE_INFINITY) return 1;
if (x == Double.NEGATIVE_INFINITY) return -1;
return x % (x - 1.d) == x ? -1 : 1;
}
public static int sign(int x) {
return Num.sign((long)x);
}
public static int sign(float x) {
return Num.sign((double)x);
}
public static void main(String args[]) {
System.out.println(Num.sign(Integer.MAX_VALUE)); // 1
System.out.println(Num.sign(1)); // 1
System.out.println(Num.sign(0)); // 0
System.out.println(Num.sign(-1)); // -1
System.out.println(Num.sign(Integer.MIN_VALUE)); // -1
System.out.println(Num.sign(Long.MAX_VALUE)); // 1
System.out.println(Num.sign(1L)); // 1
System.out.println(Num.sign(0L)); // 0
System.out.println(Num.sign(-1L)); // -1
System.out.println(Num.sign(Long.MIN_VALUE)); // -1
System.out.println(Num.sign(Double.POSITIVE_INFINITY)); // 1
System.out.println(Num.sign(Double.MAX_VALUE)); // 1
System.out.println(Num.sign(0.5d)); // 1
System.out.println(Num.sign(0.d)); // 0
System.out.println(Num.sign(-0.5d)); // -1
System.out.println(Num.sign(Double.MIN_VALUE)); // -1
System.out.println(Num.sign(Double.NEGATIVE_INFINITY)); // -1
System.out.println(Num.sign(Float.POSITIVE_INFINITY)); // 1
System.out.println(Num.sign(Float.MAX_VALUE)); // 1
System.out.println(Num.sign(0.5f)); // 1
System.out.println(Num.sign(0.f)); // 0
System.out.println(Num.sign(-0.5f)); // -1
System.out.println(Num.sign(Float.MIN_VALUE)); // -1
System.out.println(Num.sign(Float.NEGATIVE_INFINITY)); // -1
System.out.println(Num.sign(Float.NaN)); // Throws an exception
}
}
This code covers all cases and types:
public static boolean isNegative(Number number) {
return (Double.doubleToLongBits(number.doubleValue()) & Long.MIN_VALUE) == Long.MIN_VALUE;
}
This method accepts any of the wrapper classes (Integer, Long, Float and Double) and thanks to auto-boxing any of the primitive numeric types (int, long, float and double) and simply checks it the high bit, which in all types is the sign bit, is set.
It returns true when passed any of:
any negative int/Integer
any negative long/Long
any negative float/Float
any negative double/Double
Double.NEGATIVE_INFINITY
Float.NEGATIVE_INFINITY
and false otherwise.
Untested, but illustrating my idea:
boolean IsNegative<T>(T v) {
return (v & ((T)-1));
}
It seems arbitrary to me because I don't know how you would get the number as any type, but what about checking Abs(number) != number? Maybe && number != 0
Integers are trivial; this you already know. The deep problem is how to deal with floating-point values. At that point, you've got to know a bit more about how floating point values actually work.
The key is Double.doubleToLongBits(), which lets you get at the IEEE representation of the number. (The method's really a direct cast under the hood, with a bit of magic for dealing with NaN values.) Once a double has been converted to a long, you can just use 0x8000000000000000L as a mask to select the sign bit; if zero, the value is positive, and if one, it's negative.
If it is a valid answer
boolean IsNegative(char[] v) throws NullPointerException, ArrayIndexOutOfBoundException
{
return v[0]=='-';
}
one more option I could think of
private static boolean isPositive(Object numberObject) {
Long number = Long.valueOf(numberObject.toString());
return Math.sqrt((number * number)) != number;
}
private static boolean isPositive(Object numberObject) {
Long number = Long.valueOf(numberObject.toString());
long signedLeftShifteredNumber = number << 1; // Signed left shift
long unsignedRightShifterNumber = signedLeftShifteredNumber >>> 1; // Unsigned right shift
return unsignedRightShifterNumber == number;
}
This one is roughly based on ItzWarty's answer, but it runs in logn time! Caveat: Only works for integers.
Boolean isPositive(int a)
{
if(a == -1) return false;
if(a == 0) return false;
if(a == 1) return true;
return isPositive(a/2);
}
I think there is a very simple solution:
public boolean isPositive(int|float|double|long i){
return (((i-i)==0)? true : false);
}
tell me if I'm wrong!
Try this without the code: (x-SQRT(x^2))/(2*x)
Write it using the conditional then take a look at the assembly code generated.
Why not get the square root of the number? If its negative - java will throw an error and we will handle it.
try {
d = Math.sqrt(THE_NUMBER);
}
catch ( ArithmeticException e ) {
console.putln("Number is negative.");
}
I don't know how exactly Java coerces numeric values, but the answer is pretty simple, if put in pseudocode (I leave the details to you):
sign(x) := (x == 0) ? 0 : (x/x)
If you are allowed to use "==" as seems to be the case, you can do something like that taking advantage of the fact that an exception will be raised if an array index is out of bounds. The code is for double, but you can cast any numeric type to a double (here the eventual loss of precision would not be important at all).
I have added comments to explain the process (bring the value in ]-2.0; -1.0] union [1.0; 2.0[) and a small test driver as well.
class T {
public static boolean positive(double f)
{
final boolean pos0[] = {true};
final boolean posn[] = {false, true};
if (f == 0.0)
return true;
while (true) {
// If f is in ]-1.0; 1.0[, multiply it by 2 and restart.
try {
if (pos0[(int) f]) {
f *= 2.0;
continue;
}
} catch (Exception e) {
}
// If f is in ]-2.0; -1.0] U [1.0; 2.0[, return the proper answer.
try {
return posn[(int) ((f+1.5)/2)];
} catch (Exception e) {
}
// f is outside ]-2.0; 2.0[, divide by 2 and restart.
f /= 2.0;
}
}
static void check(double f)
{
System.out.println(f + " -> " + positive(f));
}
public static void main(String args[])
{
for (double i = -10.0; i <= 10.0; i++)
check(i);
check(-1e24);
check(-1e-24);
check(1e-24);
check(1e24);
}
The output is:
-10.0 -> false
-9.0 -> false
-8.0 -> false
-7.0 -> false
-6.0 -> false
-5.0 -> false
-4.0 -> false
-3.0 -> false
-2.0 -> false
-1.0 -> false
0.0 -> true
1.0 -> true
2.0 -> true
3.0 -> true
4.0 -> true
5.0 -> true
6.0 -> true
7.0 -> true
8.0 -> true
9.0 -> true
10.0 -> true
-1.0E24 -> false
-1.0E-24 -> false
1.0E-24 -> true
1.0E24 -> true
Well, taking advantage of casting (since we don't care what the actual value is) perhaps the following would work. Bear in mind that the actual implementations do not violate the API rules. I've edited this to make the method names a bit more obvious and in light of #chris' comment about the {-1,+1} problem domain. Essentially, this problem does not appear to solvable without recourse to API methods within Float or Double that reference the native bit structure of the float and double primitives.
As everybody else has said: Stupid interview question. Grr.
public class SignDemo {
public static boolean isNegative(byte x) {
return (( x >> 7 ) & 1) == 1;
}
public static boolean isNegative(short x) {
return (( x >> 15 ) & 1) == 1;
}
public static boolean isNegative(int x) {
return (( x >> 31 ) & 1) == 1;
}
public static boolean isNegative(long x) {
return (( x >> 63 ) & 1) == 1;
}
public static boolean isNegative(float x) {
return isNegative((int)x);
}
public static boolean isNegative(double x) {
return isNegative((long)x);
}
public static void main(String[] args) {
// byte
System.out.printf("Byte %b%n",isNegative((byte)1));
System.out.printf("Byte %b%n",isNegative((byte)-1));
// short
System.out.printf("Short %b%n",isNegative((short)1));
System.out.printf("Short %b%n",isNegative((short)-1));
// int
System.out.printf("Int %b%n",isNegative(1));
System.out.printf("Int %b%n",isNegative(-1));
// long
System.out.printf("Long %b%n",isNegative(1L));
System.out.printf("Long %b%n",isNegative(-1L));
// float
System.out.printf("Float %b%n",isNegative(Float.MAX_VALUE));
System.out.printf("Float %b%n",isNegative(Float.NEGATIVE_INFINITY));
// double
System.out.printf("Double %b%n",isNegative(Double.MAX_VALUE));
System.out.printf("Double %b%n",isNegative(Double.NEGATIVE_INFINITY));
// interesting cases
// This will fail because we can't get to the float bits without an API and
// casting will round to zero
System.out.printf("{-1,1} (fail) %b%n",isNegative(-0.5f));
}
}
This solution uses no conditional operators, but relies on catching two excpetions.
A division error equates to the number originally being "negative". Alternatively, the number will eventually fall off the planet and throw a StackOverFlow exception if it is positive.
public static boolean isPositive( f)
{
int x;
try {
x = 1/((int)f + 1);
return isPositive(x+1);
} catch (StackOverFlow Error e) {
return true;
} catch (Zero Division Error e) {
return false;
}
}
What about the following?
T sign(T x) {
if(x==0) return 0;
return x/Math.abs(x);
}
Should work for every type T...
Alternatively, one can define abs(x) as Math.sqrt(x*x),
and if that is also cheating, implement your own square root function...
if (((Double)calcYourDouble()).toString().contains("-"))
doThis();
else doThat();
Combined generics with double API. Guess it's a bit of cheating, but at least we need to write only one method:
static <T extends Number> boolean isNegative(T number)
{
return ((number.doubleValue() * Double.POSITIVE_INFINITY) == Double.NEGATIVE_INFINITY);
}
Two simple solutions. Works also for infinities and numbers -1 <= r <= 1
Will return "positive" for NaNs.
String positiveOrNegative(double number){
return (((int)(number/0.0))>>31 == 0)? "positive" : "negative";
}
String positiveOrNegative(double number){
return (number==0 || ((int)(number-1.0))>>31==0)? "positive" : "negative";
}
There is a function is the math library called signnum.
http://www.tutorialspoint.com/java/lang/math_signum_float.htm
http://www.tutorialspoint.com/java/lang/math_signum_double.htm
It's easy to do this like
private static boolean isNeg(T l) {
return (Math.abs(l-1)>Math.abs(l));
}
static boolean isNegative(double v) {
return new Double(v).toString().startsWith("-");
}