This question already has answers here:
How to cast a double to an int in Java by rounding it down?
(9 answers)
Closed 9 years ago.
Quick question, What is the best way to convert an primitive data type double to a primitive data type int. Currently I am using the following code but it seems to be a lot of methods to do a simple task.
double i = 2.0;
int i2 = Integer.parseInt(Double.toString(i));
just cast double to (int)
double i = 2.0;
int i2 = (int) i;
You could either cast the double to an int:
double i = 2.0
int i2 = (int) i;
or, you could round the number to the nearest whole number, then cast it to an int, just in case you would like to round:
double i = 2.2
int i2 = (int) Math.round(i);
or, you could use
double i = 2.0
int i2 = Integer.parseInt(String.valueOf(i));
About best, though, the first option would be best for the least code, and the second would be better if you would like to get the closest integer to the double.
You can turn any object from one compatible type to another by using 'type-casting'. You use this syntax:
double i = 2.0;
int i2 = (int) i;
Notice how I just put (int) before the variable i? This basically says 'hey, try to make i an int, if you can'. Since java can easily convert a float to an int, it will.
Also you can use Math.floor().
public class Example {
public static void main(String args[]) {
double i = 2.9;
System.out.println((int)(i));
System.out.println((int)Math.floor(i));
System.out.println((int)(-i));
System.out.println((int)Math.floor(-i));
}
}
output:
2
2
-2
-3
Related
This question already has answers here:
convert double into int
(1 answer)
Cast Double to Integer in Java
(19 answers)
Closed 5 years ago.
A project I have requires the movement of a player at a coordinate using getX and getY, however, I am confused on how to convert getX and getY from a double to an int so i can play them in the drawing panel.
Just casting to int will truncate the double. So you need to specify what result you really want before you decide how to get the int. For example, if the double value is 2.999, do you want your int to be 2 or 3?
If you want the closest int (3 above), then use Math.round(d) which returns a long.
You want to cast your double as an int but you need to be careful about how you cast because if you want your movement to be accurate you want values of x.5 and over to be rounded up and values below x.5 to be rounded down.
Casting will always round down so a good way to properly round is to add .5 to all of your doubles before you cast to an int.
Here are a few examples
double = 1.1
int (double) = 1
double = 1.7
int (double) = 1 \\ Note that we will likely want this to be 2
Using our method lets see how these 2 doubles would be casted
double = 1.1
double + .5 = 1.6
int (double) = 1
double 1.7 = 1.7
double + .5 = 2.2
int (double) = 2
Note that now our doubles that are above x.5 will be rounded up properly.
This question already has answers here:
How do I get the part after the decimal point in Java?
(15 answers)
Closed 6 years ago.
I am developing an app in which i have to get the float part in terms of integer value.
for example if my number is 153.12324 then output should be 12324.I tried this way but it works wrong some time.
What's the problem in this or is there a better way to do this ?
there is a double value in d
double d =any_double_value;
String[] splitter = String.valueOf(d).split("\\.");
splitter[0].length(); // Before Decimal Count
int count = splitter[1].length();
int integerpart = (int) calcResult;
double floatpart = calcResult - integerpart;
while (count!=0)
{
floatpart=floatpart*10;
count--;
}
int floatpartinInt=(int)floatpart;
this works wrong only in some cases(it gives 1 number less, like if answer is 124 it gives 123) and in cases where answer is long double no like 3.333333333 (10/3)
You can also without split. by using following way,
double value = 3.25;
double fractionalPart = value % 1;
double integralPart = value - fractionalPart;
This may helps you
Use this code its work in your case and enjoy Man
double val=1.9;
String[] arr=String.valueOf(val).split("\\.");
int[] intArr=new int[2];
intArr[0]=Integer.parseInt(arr[0]); // 1
intArr[1]=Integer.parseInt(arr[1]); // 9
You could also do this:
lvalue = (long) value;
fPart = value - lvalue;
This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Closed 6 years ago.
The question require me to write a Java program to show the results of the following cast operator expressions:
(double) (23 / 14) + 7.65
My Code:
public class op {
public static void main(String [] args) {
int num = 23/14;
double r1 = (double) num;
double result = r1 + 7.65;
System.out.println("Results: "+ result);
}
}
I don't think I have done correctly, what are the problems of my code?
By the way, can someone tell me what are the differences between long, double, int, float? How do we know when to use these primitive data types? I read an explanation here: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
but is there any 'human-version' of the explanation?
Thank you for your help.
The problem is due to the used types.
Since you divide two integers (23 and 14), the result is considered and int as well. Therefor, 23/14 = 1.642857142857143, which is truncated to fit in an int result, more specifically, 1.
result is the sum of 1 (int) and 7.65 (double). Since one of them is a double, to other is converted to the "upper" type as well (double) and the operation becomes 1.0+7.65 = 8.65.
The result is correct, because you asked the result of (double) (23 / 14) + 7.65 which means the result of casting the result of the operations in brackets to double summed with 7.65. Which is 8.65 as previously explained.
If you want to use a division using doubles, consider:
double r1 = 1.0 * 23/14;
Lets see step-by-step:
int num = 23/14; // int division of 23/14 results in 1
So, here num = 1
When you cast num to double value of r1 is setted to 1.0.
double result = r1 + 7.65; //1.0 + 7.65 = 8.65
ok, int is short term for INTEGER which are natural numbers that we use normally but with no decimal places and if your number has some value in between roughly -2 billion to +2 billion. if your range exceeds that and you still want an integer then go for long data type.
floats are for decimal values like 3.147 with a range of +10*38 to -10*38 or so, but if your range exceeds this(practically this happens rarely) go for double.
coming to the code you put here , if you divide a int by another int (like 23/14) you get only get the integer part of the answer(only '1' in 23/14=1.642...) , next when you cast it to double you get 1.0 and next you are going to add that to 7.65 which will make the ultimate answer as 8.65 hope this answers your Q....
You could change this int num = 23/14
to double num = ((double) 23)/14
or double num = (23 * 1.0)/14
I've got an excercise from university which looks like:
int a = 10;
int b = 3;
double c = a / b;
The question is: Which value is c.
Now I would say, c is 3.3. It's casted implicit to double before calculating the result.
But the correct answer to this question is according to my records 3.0.
How this can be? Does the compiler really calculate the result first as integer and then in a second step casts it to double?
Or did I understand that incorrectly?
Does the compiler really calculate the result first as integer and
then in a second step casts it to double?
Yes
Does the compiler really calculate the result first as integer and
then in a second step casts it to double?
Yes,
Runtime first calculates the RHS result and then converts the result to double. Now in your case as RHS contains int / int so the result is in int and you don't get 3.3.
So if RHS contains double / int or int / double, the type promotion occurs and RHS operands are promoted to double before calculating the result and hence you get 3.3
See what is actually happening is :
double c = (double) a / b; //double of 3 = 3.0
you have to do
double c = a/(double)b
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Dividing two integers in Java gives me 0 or 100?
(5 answers)
Closed 9 years ago.
OK so this is a problem in a larger piece of code which does'nt seem to make sense. below is the code which is the problem... It prints Ratio = 0.0 the console when launched it should be equal to ~0,348.
public class MathTest {
public static void main(String[] args) {
double ratio = 29511 / 84812;
System.out.println("Ratio = "+ ratio);
}
}
Edit:
What if the code is this:
public class MathTest {
public static void main(String[] args) {
int int1 = 7;
int int2 = 13;
double double1 = int1/int2;
System.out.println("double1 = "+ double1);
}
}
It again prints "0.0".
You have unwittingly used integer division when you say 29511 / 84812, which in Java, loses the decimal points. Use double literals (with .0 added) instead:
double ratio = 29511.0 / 84812.0;
Other solutions that work here:
Cast one of them to a double: (double) 29511 / 84812
Use 'D' as another way to indicate a Java double literal: 29511D / 84812D
You are doing integer division because 29511 and 84812 are ints. The result of the division is 0, and you are saving it as a double so it becomes 0.0.
To fix this, cast one of the operands on the right side of the assignment to a double.
double ratio = (double)29511 / 84812;
Replace double ratio = 29511 / 84812 with double ratio = 29511.0 / 84812.0;
It make perfect sense. You are dividing two integers to get an integer. This means 0 in this case.