Develop Reference

Why is this HackerRank problem working fine with 'int' but not with 'Integer'? [duplicate]

I know that if you compare a boxed primitive Integer with a constant such as:
Integer a = 4;
if (a < 5)
a will automatically be unboxed and the comparison will work.
However, what happens when you are comparing two boxed Integers and want to compare either equality or less than/greater than?
Integer a = 4;
Integer b = 5;
if (a == b)
Will the above code result in checking to see if they are the same object, or will it auto-unbox in that case?
What about:
Integer a = 4;
Integer b = 5;
if (a < b)
?

No, == between Integer, Long etc will check for reference equality - i.e.
Integer x = ...;
Integer y = ...;
System.out.println(x == y);
this will check whether x and y refer to the same object rather than equal objects.
So
Integer x = new Integer(10);
Integer y = new Integer(10);
System.out.println(x == y);
is guaranteed to print false. Interning of "small" autoboxed values can lead to tricky results:
Integer x = 10;
Integer y = 10;
System.out.println(x == y);
This will print true, due to the rules of boxing (JLS section 5.1.7). It's still reference equality being used, but the references genuinely are equal.
If the value p being boxed is an integer literal of type int between
-128 and 127 inclusive (§3.10.1), or the boolean literal true or false (§3.10.3), or a character literal between '\u0000' and '\u007f'
inclusive (§3.10.4), then let a and b be the results of any two boxing
conversions of p. It is always the case that a == b.
Personally I'd use:
if (x.intValue() == y.intValue())
or
if (x.equals(y))
As you say, for any comparison between a wrapper type (Integer, Long etc) and a numeric type (int, long etc) the wrapper type value is unboxed and the test is applied to the primitive values involved.
This occurs as part of binary numeric promotion (JLS section 5.6.2). Look at each individual operator's documentation to see whether it's applied. For example, from the docs for == and != (JLS 15.21.1):
If the operands of an equality
operator are both of numeric type, or
one is of numeric type and the other
is convertible (§5.1.8) to numeric
type, binary numeric promotion is
performed on the operands (§5.6.2).
and for <, <=, > and >= (JLS 15.20.1)
The type of each of the operands of a
numerical comparison operator must be
a type that is convertible (§5.1.8) to
a primitive numeric type, or a
compile-time error occurs. Binary
numeric promotion is performed on the
operands (§5.6.2). If the promoted
type of the operands is int or long,
then signed integer comparison is
performed; if this promoted type is
float or double, then floating-point
comparison is performed.
Note how none of this is considered as part of the situation where neither type is a numeric type.

== will still test object equality. It is easy to be fooled, however:
Integer a = 10;
Integer b = 10;
System.out.println(a == b); //prints true
Integer c = new Integer(10);
Integer d = new Integer(10);
System.out.println(c == d); //prints false
Your examples with inequalities will work since they are not defined on Objects. However, with the == comparison, object equality will still be checked. In this case, when you initialize the objects from a boxed primitive, the same object is used (for both a and b). This is an okay optimization since the primitive box classes are immutable.

Since Java 1.7 you can use Objects.equals:
java.util.Objects.equals(oneInteger, anotherInteger);
Returns true if the arguments are equal to each other and false
otherwise. Consequently, if both arguments are null, true is returned
and if exactly one argument is null, false is returned. Otherwise,
equality is determined by using the equals method of the first
argument.

We should always go for the equals() method for comparison of two integers. It's the recommended practice.
If we compare two integers using == that would work for certain range of integer values (Integer from -128 to 127) due to the JVM's internal optimisation.
Please see examples:
Case 1:
Integer a = 100;
Integer b = 100;
if (a == b) {
System.out.println("a and b are equal");
} else {
System.out.println("a and b are not equal");
}
In above case JVM uses value of a and b from cached pool and return the same object instance(therefore memory address) of integer object and we get both are equal.Its an optimisation JVM does for certain range values.
Case 2: In this case, a and b are not equal because it does not come with the range from -128 to 127.
Integer a = 220;
Integer b = 220;
if (a == b) {
System.out.println("a and b are equal");
} else {
System.out.println("a and b are not equal");
}
Proper way:
Integer a = 200;
Integer b = 200;
System.out.println("a == b? " + a.equals(b)); // true

tl;dr my opinion is to use a unary + to trigger the unboxing on one of the operands when checking for value equality, and simply use the maths operators otherwise. Rationale follows:
It has been mentioned already that == comparison for Integer is identity comparison, which is usually not what a programmer want, and that the aim is to do value comparison; still, I've done a little science about how to do that comparison most efficiently, both in term of code compactness, correctness and speed.
I used the usual bunch of methods:
public boolean method1() {
Integer i1 = 7, i2 = 5;
return i1.equals( i2 );
}
public boolean method2() {
Integer i1 = 7, i2 = 5;
return i1.intValue() == i2.intValue();
}
public boolean method3() {
Integer i1 = 7, i2 = 5;
return i1.intValue() == i2;
}
public boolean method4() {
Integer i1 = 7, i2 = 5;
return i1 == +i2;
}
public boolean method5() { // obviously not what we want..
Integer i1 = 7, i2 = 5;
return i1 == i2;
}
and got this code after compilation and decompilation:
public boolean method1() {
Integer var1 = Integer.valueOf( 7 );
Integer var2 = Integer.valueOf( 5 );
return var1.equals( var2 );
}
public boolean method2() {
Integer var1 = Integer.valueOf( 7 );
Integer var2 = Integer.valueOf( 5 );
if ( var2.intValue() == var1.intValue() ) {
return true;
} else {
return false;
}
}
public boolean method3() {
Integer var1 = Integer.valueOf( 7 );
Integer var2 = Integer.valueOf( 5 );
if ( var2.intValue() == var1.intValue() ) {
return true;
} else {
return false;
}
}
public boolean method4() {
Integer var1 = Integer.valueOf( 7 );
Integer var2 = Integer.valueOf( 5 );
if ( var2.intValue() == var1.intValue() ) {
return true;
} else {
return false;
}
}
public boolean method5() {
Integer var1 = Integer.valueOf( 7 );
Integer var2 = Integer.valueOf( 5 );
if ( var2 == var1 ) {
return true;
} else {
return false;
}
}
As you can easily see, method 1 calls Integer.equals() (obviously), methods 2-4 result in exactly the same code, unwrapping the values by means of .intValue() and then comparing them directly, and method 5 just triggers an identity comparison, being the incorrect way to compare values.
Since (as already mentioned by e.g. JS) equals() incurs an overhead (it has to do instanceof and an unchecked cast), methods 2-4 will work with exactly the same speed, noticingly better than method 1 when used in tight loops, since HotSpot is not likely to optimize out the casts & instanceof.
It's quite similar with other comparison operators (e.g. </>) - they will trigger unboxing, while using compareTo() won't - but this time, the operation is highly optimizable by HS since intValue() is just a getter method (prime candidate to being optimized out).
In my opinion, the seldom used version 4 is the most concise way - every seasoned C/Java developer knows that unary plus is in most cases equal to cast to int/.intValue() - while it may be a little WTF moment for some (mostly those who didn't use unary plus in their lifetime), it arguably shows the intent most clearly and most tersely - it shows that we want an int value of one of the operands, forcing the other value to unbox as well. It is also unarguably most similar to the regular i1 == i2 comparison used for primitive int values.
My vote goes for i1 == +i2 & i1 > i2 style for Integer objects, both for performance & consistency reasons. It also makes the code portable to primitives without changing anything other than the type declaration. Using named methods seems like introducing semantic noise to me, similar to the much-criticized bigInt.add(10).multiply(-3) style.

== checks for reference equality, however when writing code like:
Integer a = 1;
Integer b = 1;
Java is smart enough to reuse the same immutable for a and b, so this is true: a == b. Curious, I wrote a small example to show where java stops optimizing in this way:
public class BoxingLol {
public static void main(String[] args) {
for (int i = 0; i < Integer.MAX_VALUE; i++) {
Integer a = i;
Integer b = i;
if (a != b) {
System.out.println("Done: " + i);
System.exit(0);
}
}
System.out.println("Done, all values equal");
}
}
When I compile and run this (on my machine), I get:
Done: 128

Calling
if (a == b)
Will work most of the time, but it's not guaranteed to always work, so do not use it.
The most proper way to compare two Integer classes for equality, assuming they are named 'a' and 'b' is to call:
if(a != null && a.equals(b)) {
System.out.println("They are equal");
}
You can also use this way which is slightly faster.
if(a != null && b != null && (a.intValue() == b.intValue())) {
System.out.println("They are equal");
}
On my machine 99 billion operations took 47 seconds using the first method, and 46 seconds using the second method. You would need to be comparing billions of values to see any difference.
Note that 'a' may be null since it's an Object. Comparing in this way will not cause a null pointer exception.
For comparing greater and less than, use
if (a != null && b!=null) {
int compareValue = a.compareTo(b);
if (compareValue > 0) {
System.out.println("a is greater than b");
} else if (compareValue < 0) {
System.out.println("b is greater than a");
} else {
System.out.println("a and b are equal");
}
} else {
System.out.println("a or b is null, cannot compare");
}

In my case I had to compare two Integers for equality where both of them could be null. I searched similar topics, but I didn't find anything elegant for this. I came up with simple utility function:
public static boolean integersEqual(Integer i1, Integer i2) {
if (i1 == null && i2 == null) {
return true;
}
if (i1 == null && i2 != null) {
return false;
}
if (i1 != null && i2 == null) {
return false;
}
return i1.intValue() == i2.intValue();
}
// Considering null is less than not-null
public static int integersCompare(Integer i1, Integer i2) {
if (i1 == null && i2 == null) {
return 0;
}
if (i1 == null && i2 != null) {
return -1;
}
return i1.compareTo(i2);
}

Because a comparison method has to be done based on type int (x==y) or class Integer (x.equals(y)) with the right operator:
public class Example {
public static void main(String[] args) {
int[] arr = {-32735, -32735, -32700, -32645, -32645, -32560, -32560};
for(int j=1; j<arr.length-1; j++)
if((arr[j-1] != arr[j]) && (arr[j] != arr[j+1]))
System.out.println("int>" + arr[j]);
Integer[] I_arr = {-32735, -32735, -32700, -32645, -32645, -32560, -32560};
for(int j=1; j<I_arr.length-1; j++)
if((!I_arr[j-1].equals(I_arr[j])) && (!I_arr[j].equals(I_arr[j+1])))
System.out.println("Interger>" + I_arr[j]);
}
}

This method compares two Integer's with a null check. See the tests.
public static boolean compare(Integer int1, Integer int2) {
if(int1!=null) {
return int1.equals(int2);
} else {
return int2==null;
}
//inline version:
//return (int1!=null) ? int1.equals(int2) : int2==null;
}
//results:
System.out.println(compare(1,1)); //true
System.out.println(compare(0,1)); //false
System.out.println(compare(1,0)); //false
System.out.println(compare(null,0)); //false
System.out.println(compare(0,null)); //false
System.out.println(compare(null,null)); //true

why 2 objects of Integer class in Java cannot be equal

my code is:
public class Box
{
public static void main(String[] args)
{
Integer z = new Integer(43);
z++;
Integer h = new Integer(44);
System.out.println("z == h -> " + (h == z ));
}
}
Output:-
z == h -> false
why the output is false when the values of both the objects is equal?
Is there any other way in which we can make the objects equal?

No. Use h.equals(z) instead of h == z to get the equality behavior you expect.

h == z would work only if you assign the value via auto-boxing (i.e Integer a = 43) and the value is in between -128 and 127 (cached values), i.e:
Integer a = 44;
Integer b = 44;
System.out.println("a == b -> " + (a == b));
OUTPUT:
a == b -> true
If the value is out of the range [-128, 127], then it returns false
Integer a = 1000;
Integer b = 1000;
System.out.println("a == b -> " + (a == b));
OUTPUT:
a == b -> false
However, the right way to compare two objects is to use Integer.equals() method.

Integer is Object not primitive (int) And Object equality compare with equals method.
When you do z == h it will not unbox into int value unless it checks both Integer reference(z & h) are referring same reference or not.
As it is derived in documentation -
The result is true if and only if the argument is not null and is an
Integer object that contains the same int value as this object.
System.out.println("z == h -> " + h.equals( z));
It will print true.

You are trying to compare two different object and not their values. z and h points to two different Integer object which hold same value.
z == h
Will check if two objects are equal. So it will return false.
If you want to compare values stored by Integer object use equals method.
Integer z = new Integer(43); // Object1 is created with value as 43.
z++; // Now object1 holds 44.
Integer h = new Integer(44); // Object2 is created with value as 44.
So at the end we have two different Integer object ie object1 and object2 with value as 44.
Now
z = h
This will check if objects pointed by z and h is same. ie object1 == object2
which is false.
If you do
Integer z = new Integer(43); // Object1 is created with value as 43.
z++; // Now object1 holds 44. Z pointing to Object1
Integer h = z; // Now h is pointing to same object as z.
Now
z == h
will return true.
This might help http://www.programmerinterview.com/index.php/java-questions/java-whats-the-difference-between-equals-and/

Integer is an object, not a primitive. If z & h were primitives, == would work just fine. When dealing with objects, the == operator doesn't check for equality; it checks if the two references point to the same object.
As such, use z.equals(h); or h.equals(z); These should return true.

Read this: http://docs.oracle.com/javase/1.5.0/docs/api/java/lang/Integer.html
Integer is Object you compare address/references/pointers of objects not values.
Integer a = Integer(1);
Integer b = Integer(1);
a == b; // false
a.compareTo(b); // true

check to make sure you can use z++ on the Integer object.

How to test if a double is an integer

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
}

Using the == operator twice in a if statement

Is it okay to do like this in java, does it work?
if (turtles.get(h).getX() == turtles.get(g).getX() == 450) {
//stuff here
}
Basically, i want to check if X is the same value as Y and that value should be 450.

No. What do you expect to happen there?
"a == b" evaluates into a boolean, so "int == (int == int)" would evaluate into "int == boolean", and you cannot compare and int and a boolean.
Besides, what kind of logic are you trying to do here? if ((a == b) && (b == c))?

No, it's not. This is because the result of a == b is a boolean. If you do a == b == c you are first comparing a == b which will return true or false and then comparing that truth value to c.
Not what you want to do, usually!
Note that this trick can work for assignment because the result of a = b is b (the new value of a) which means a = b = c or even (a = b) == c come in useful occasionally.

No. It is the same as (turtles.get(h).getX() == turtles.get(g).getX()) == 450 - "incomparable types". if(turtles.get(h).getX() == 450 && turtles.get(g).getX() == 450).

Or avoid all the less-readable (and error-prone) repetition with a helper method...
public boolean areEqual( int a, int b, int c )
{
return ( a == b ) && ( b == c ) ;
}

That won't work, because the == operator is binary.
And even if it worked sequentially, the first set would return a boolean, which won't work against the integer that follows.

No it won't work, as explained in the other posts. But you could do
if (turtles.get(h).getX() - turtles.get(g).getX() + 450 == 0)

How to determine if a number is positive or negative?

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("-");
}