How do I convert/cast a Double object to an int?
Double d = new Double(12.34);
int i = 12 //is what I'm looking for
If I had a double I would just use:
double z = 12.34;
int i = (int)z;
Note: I want it to truncate the Double (so 12.9 becomes 12) because I know that d actually was an integer that was converted to a Double.
I.e. I want to call (int) on a Double.
To perform this operation you need to call the intValue method on the Double object:
int i = d.intValue();
This method performs a cast, into an int value, on the double value wrapped within the Double object, d.
Note that it does not check for null before attempting this.
If you want to use a cast you can use
Double d = 12.34;
int i = (int) (double) d;
You can't cast from Double to int in one step.
I have a simple function as defined below:
if(currentValue < smallestValue)
{
smallestValue = currentValue;
}
CurrentValue can be any value of int.
All my values are ints
What can I initialize smallestValue to so it will always be greater than the first currentValue?
By this rule: CurrentValue can be any real number, use Double.MAX_VALUE:
double smallestValue = Double.MAX_VALUE;
But the second rule negates the previous: All my values are ints, so use Integer.MAX_VALUE
int smallestValue = Integer.MAX_VALUE;
I'm making a simple calculator (I'm a beginner), and I was wondering why when I divide 1 by 4, I get 0. I know it has something to do with the type of number. Here is my code:
private static void calc(int a, int b,String op){
if (op.equals("add")){
double ans = a+b;
System.out.println(ans);
}
if(op.equals("subtract")){
double ans=a-b;
System.out.println(ans);
}
if(op.equals("multiply")){
double ans=a*b;
System.out.println(ans);
}
if(op.equals("divide")){
double ans=a/b;
System.out.println(ans);
}
}
I can't get my variable (ans) to have more than one decimal spot. And when the answer requires one (like 1/4), it just returns 0.
Help please.
You need to divide on a double since all your result values in double.
private static void calc(int a, double b,String op){
}
Or simply :
if(op.equals("divide")){
double ans=(double)a/b;
System.out.println(ans);
}
Although the other comments and answers are correct, they do not give enough detail (IMO).
Although ans is a double, the value of the right side of the equation is still an int.
double ans = a / b;
// equiv to:
// double ans = (double)(a / b);
So when you evaluate this you get:
double ans = (double)(1/4);
= (double)(0);
= 0.0d;
To get around this, you will need to cast one or both of the operands to a double.
double ans = ( (double) a ) / ( (double) b );
= ( (double) 1 ) / ( (double) 4 );
= 1.0d / 4.0d;
~= 0.25d;
Also note with regards to #Salah's answer that the cast from integer to double has just been moved. Since you can implicitly convert an int to a double, calling the function with an integer (such as the literal 1) will automatically convert the integer to a double. However, you may not want to allow non-integral numbers to be passed to your function, and if so, then this solution would not be correct.
I am going through the following code in Java:
public float a1 = 0.10f;
int Increments;
if(TotalTime > 0)
{
Increments = (int) ceil(TotalTime / Increment1);
amount = round(Increments * a1, 4);
}
Where,
TotalTime and Increment1 are integers
My questions is, Why do we need to typecast to integer just before ceil as done in the following line of code above?
Increments = (int) ceil(TotalTime / Increment1);
Because Math.ceil (I suppose it's a static import there) returns a double. As you make a restriction of the converted value (by converting to a less precise type), you need to do the conversion explicitly.
Also, please note that the argument you pass to ceil is the ratio of two integers, which is also an integer. So you're in danger of losing precision. Moreover, applying ceil to an integer is redundant, as it will return that integer itself. And the last remark: ceil expects a double as its argument. In this case, you pass an int, but the conversion is done implicitly, because it is done from a less precise to a more precise type.
The reason is that Math.ceil returns a double. You cannot implicitly cast from a double to an int. Trying to compile this:
double d = 2.1;
int i = d;
Gives you:
dur.java:4: possible loss of precision
found : double
required: int
int i = d;
^
1 error
As the error message states, the reason is that you might lose precision, so the compiler wants to be sure that you really want to allow this loss. You do this by making the cast explicit:
int i = (int)d;
Note that even so, your code won't do what you expect.
int i = 10;
int j = 3;
int k = (int)Math.ceil(i/j);
System.out.println(k); //outputs '3'
The reason is that i / j performs integer division if both i and j are integers, which rounds down. i / j is already 3, and all Math.ceil does is return 3.0, which then again gets casted to 3. You want to do floating point division instead, which is done when one or both of i and j are floating point types. You can achieve this by casting one of the integers to a double before doing the division:
int i = 10;
int j = 3;
int k = (int)Math.ceil((double)i/j);
System.out.println(k); //outputs '4'
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
}