## How to convert number from double to int without rounding?

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I want to calculate how many columns can fit into container. First I calculate how many columns can fit in, then I calculate number of margins for them. After that I check if they will fit in with margins, if no - reduce amount of columns by one. Code:

int columnMargin = 50; double result = columnContainerHBox/minimalColumnWidth; int columnCount = (int) result; // here's the problem int numberOfMargins = columnCount - 1; double finalCount = columnContainerHBox/minimalColumnWidth*columnCount+columnMargin*numberOfMargins; if(finalCount<1){ columnCount--; }

The problem is that I don't know how to convert from `double`

to `int`

without rounding number. I just need to get rid of all numbers after decimal. I already tried `(int) double`

, `(int)Math.floor(double)`

and `new DecimalFormat('#').format(double)`

. As a `double`

I took 1.99999999999999999999. All above was converting it to 2. Desirable result - 1;

My `double`

is going to be completely random. It can be 1.0, 1.9999999999, 2.45, 3.5959547324 etc.

What I need is to get a whole part and completely discard everything after decimal.

You do it by casting to int, as you're already doing in the code you've presented as an attempt.

I understand that you feel unsatisfied because you think you have found a case when your attempt will round the value instead of dropping the decimal part.

You have tried with value 1.99999999999999999999 and you've noticed that casting it to int produces an int of value 2. You concluded from there that casting is not just dropping the decimal part, it is rounding to the closest whole number 2.

Your conclusion is incorrect. The number 2 is not obtained as a result of casting to int. The number 2 is a result of the compiler parsing the written literal value 1.99999999999999999999. doubles don't have infinite precision. That means you can't write as many decimals as you want and expect it to be correctly kept in a double value. doubles only offer **approximations** of what you're asking. So when you type the literal value 1.99999999999999999999, the compiler knows that it is incapable to represent that value exactly, and instead it will take it as the **closest** value that can be represented in a double.

And the representable value closest to 1.99999999999999999999 is 2. As far as the Java compiler is concerned, these two numbers are one and the same.

So when you write:

double d = 1.99999999999999999999d;

the Java compiler treats it completely equivalent to:

double d = 2d;

You'll notice that at this point, you have yet to do any attempt to drop the decimals and only keep the whole part. You've only declared a value for your double, and as far as you're concerned this value could very well have a decimal part.

Your attempt to only keep the whole value only happens when you do:

int i = (int)d;

and here the decimals are dropped and your int contains only the whole part of what the double value contained.

However, in your example, since your double value was 2.0, then taking the whole part of it is 2. That's not rounding. That's not anything else than keeping only the whole part.

The correct way to drop decimals and only keep the whole part, is to cast to int.

(If it is important to you to be able to manipulate values such as 1.99999999999999999999 and have them not be the same as 2.0, then you cannot use doubles. They don't have sufficient precision. You should use BigDecimal, with new BigDecimal("1.99999999999999999999"). The constructor must be called with a String rather than a floating-point value, since floating-point values are unable to represent the value you want.)

**How to convert number from double to int without rounding?,** You do it by casting to int, as you're already doing in the code you've presented as an attempt. I understand that you feel unsatisfied because If our double value is within the int range, we can cast it to an int. The cast truncates the decimal part, meaning that it cuts it off without doing any rounding. This approach is about 10 times as fast as the other approaches we'll look at. Once it's an int, then we can then pass it to the valueOf method on the String class:

`Math.floor()`

is what you are looking for https://docs.oracle.com/javase/8/docs/api/java/lang/Math.html#floor-double-

`Math.floorDiv()`

also might be useful in your case.

**Java - Convert double to int example,** round() method. This method rounds of the number to nearest integer. As you can see in the output that the double value 99.99 is rounded of to the nearest If we wish to convert a Java double to int with rounding, the Math utility class has the static method round to help in this case. Math.round accepts a double value to round it to the next whole number. But since double has a big range, the result is a long value. We can just cast the resulting long value to an integer.

I have do this but it isn't a very good solution but I have that:

public class Main { public static void main(String args[]) { double d = 1.9999; String s = (""+d).contains(".") ? (""+d).substring(0, (""+d).indexOf(".")) : (""+d); System.out.println(Integer.parseInt(s)); } }

**Convert Double to Integer in Java,** round(): This method returns the nearest integer. Syntax: double data = 3452.645 int value = (int)Math.round(data);. Example: filter_none. I have a variable declared as Single and I am trying to retrieve only the whole number without rounding up or down as in the following number: 3.959 I only want to assign 3 in the number above to another variable. If I do this in code using format number and read only the whole number of course I get the value of 4.

**https://social.msdn.microsoft.com/Forums/vstudio/e,** Cast, int. Floating point values can be cast to ints. This has predictable results. First, when you convert doubles to ints, the values after the decimal place will to integers, you will know they are always rounded down to the nearest integer. Convert number without rounding – Learn more on the SQLServerCentral forums. or int. When function is omitted or has a value of 0 (default), numeric_expression is rounded.

**C# Cast to Int (Convert Double to Int),** Out of the three ways there is one way (Math.round() function) by which you can round of to the nearest integer value i.e int value will be 10 for double value 9.99. Double Rounding Errors in Floating-Point Conversions By Rick Regan August 11th, 2010 Double rounding is when a number is rounded twice, first from n 0 digits to n 1 digits, and then from n 1 digits to n 2 digits.

**3 ways : Convert double to int in java with example,** This is because no round or any operation is done. If we wish to convert a Java double to int with rounding, the Math utility class has the static method round to help in We can just cast the resulting long value to an integer. In this quick tutorial, we covered different techniques for rounding numbers to n decimal places. We can simply format the output without changing the value, or we can round the variable by using a helper method. We've also covered a few libraries that deal with this problem. The code used during the discussion can be found over on GitHub.

##### Comments

- are you looking for Floor method? if so, java.lang.Math.floor is your answer
`1.99999999999999999999`

is equal to 2 in double precision. Floating point numbers do not have perfect accuracy.- Try using Math.floor() without casting the result to an int.
- Use BigDecimal instead of double. Or round by java.lang.Math.floor
- xoxel, result will be 2.0
- does it mean that I will never run into this situation by dividing two
`double`

s? - You shouldn't run into that.
- @Alyona what situation? Casting to int will leave you with the integer part, without doing any rounding. If you have
`double d;`

provided d<2.0 and d>1.0 casting to int will leave you with 1. `System.out.println(Math.floor(1.99999999999999999999));`

prints`2.0`

, and casting it to an`int`

prints`2`

.- Strange. Looking into it.
- As @khelwood pointed out, that long of a number will be treated as 2 due to the way double precision works. You can try 1.999999999999999 and it works. Based on your real life situation, it's impossible you would get a number with your example based on dividing two integers.
- I'm going to be dividing
`double`

s - You can't create a number beyond double precision when dividing two doubles. It will round to the limits of a double.