This question already has answers here:
Why does the division of two integers return 0.0 in Java? [duplicate]
(6 answers)
Closed 8 years ago.
In the calculation of double numbers, what's the difference between, say, 6.0 and 6?
Because when I was solving a problem on online judge, the expression
estimatedPI = Math.sqrt(6*a/b);
got "Wrong answer" on OJ, while
estimatedPI = Math.sqrt(6.0*a/b);
got "Accepted" on OJ.
For the output, because I used
String result;
result = String.format("%.6f\n",estimatedPI);
System.out.print(result);
so the output looks exactly the same, with six digits after decimal point.
The estimatedPI is declared double and a,b declared int.
So why 6.0 got "Accepted" and 6 got "Wrong answer"? What would be the difference here?
Thanks.
Edit: Noted of duplicated questions.
6.0 is a double. 6 is an int.
If a and b are also ints, then 6*a/b is not a "double calculation" - it will be done using int arithmetic.
When you mix doubles and ints in a binary mathematical operation, the int is converted to a double and then the operation is done using double arithmetic. So 6.0*a does a double multiplication (converting a to double first), resulting in a double. Then (the result of that)/b also does a double division (converting b to double first).
The .0 is an indicator to the compiler that the constant is a floating point number rather than an integer. Your expression 6 * a / b will be treated as an integer expression.
Related
This question already has answers here:
Rounding Errors?
(9 answers)
Is floating point math broken?
(31 answers)
Closed 6 years ago.
Today I find an interesting fact that the formula will influence the precision of the result. Please see the below code
double x = 7d
double y = 10d
println(1-x/y)
println((y-x)/y)
I wrote this code using groovy, you can just treat it as Java
The result is
1-x/y: 0.30000000000000004
(y-x)/y: 0.3
It's interesting that the two formulas which should be equal have different result.
Can anyone explain it for me?
And can I apply the second formula to anywhere applicable as a valid solution for double precision issue?
To control the precision of floating point arithmetic, you should use java.math.BigDecimal.
You can do something like this.
BigDecimal xBigdecimal = BigDecimal.valueOf(7d);
BigDecimal yBigdecimal = BigDecimal.valueOf(10d);
System.out.println(BigDecimal.valueOf(1).subtract(xBigdecimal.divide(yBigdecimal)));
Can anyone explain it for me?
The float and double primitive types in Java are floating point numbers, where the number is stored as a binary representation of a fraction and a exponent.
More specifically, a double-precision floating point value such as the double type is a 64-bit value, where:
1 bit denotes the sign (positive or negative).
11 bits for the exponent.
52 bits for the significant digits (the fractional part as a binary).
These parts are combined to produce a double representation of a value.
Check this
For a detailed description of how floating point values are handled in Java, follow Floating-Point Types, Formats, and Values of the Java Language Specification.
The byte, char, int, long types are fixed-point numbers, which are exact representions of numbers. Unlike fixed point numbers, floating point numbers will some times (safe to assume "most of the time") not be able to return an exact representation of a number. This is the reason why you end up with 0.30000000000000004 in the result of 1 - (x / y).
When requiring a value that is exact, such as 1.5 or 150.1005, you'll want to use one of the fixed-point types, which will be able to represent the number exactly.
As I've already showed in the above example, Java has a BigDecimal class which will handle very large numbers and very small numbers.
This question already has answers here:
Division of integers in Java [duplicate]
(7 answers)
Closed 6 years ago.
This feels like a stupid question, but I can't find the answer anywhere in the Java documentation. If I declare two ints and then divide them, what exactly is happening? Are they converted to floats/doubles first, divided, then cast back to an integer, or is the division "done" as integers?
Also, purely from experimentation, integer division seems to round the answer towards zero (i.e. 3/2 = 1 and -3/2 = -1). Am I right in believing this?
They are being divided in integer arithmetics. So dividing integer a by integer b you get how many times b fits into a. Also a % b will give you a remainder of a division. So (a / b ) * b + a % b = a
Java does autoconvert types:
"It autoconverts ints to doubles. It autoconverts shorts and bytes to ints even when no ints are involved, requiring constant annoying casts when you want to do short or byte arithmetic. It autoconverts primitives to wrappers and vice versa for boxing and autoboxing." - user2357112
Java never casts anything without you specifying it.
But still integer / integer = integer.
Also, it does always truncate the result. So if the result would be 0.999999 as float the integer division would still return 0.
This question already has answers here:
Why not use Double or Float to represent currency?
(16 answers)
Is floating point math broken?
(31 answers)
Closed 7 years ago.
I am working on a project where we are calculating prices of items with 8 places of decimal points.
But sometime calculation results are different.
I tried some samples but getting different calculation results:
public static void main(String[] args) {
double value1 = 0.999939f * 0.987792f;
double value2 = 0.999939 * 0.987792;
double value3 = (double)0.999939f * 0.987792f;
System.out.println("value 1 : "+ value1);
System.out.println("value 2: "+ value2);
System.out.println("value 3 : "+ value3);
}
Outputs are:
value 1 : 0.9877317547798157
value 2 : 0.9877317446880001
value 3 : 0.9877317839126931
These three are different results.
I am confused. Can anyone please clarify me what is happening here?
Thanks.
I went through some already answered, But I just want some ways where I can calculate with float and doubles only. otherwise I have to change many places. It will be painful.
Because of how floats and decimals are represented, the casts could be causing different results.
For: double value1 = 0.999939f * 0.987792f you are multiplying two floats and then casting it to a double. The resulting float representation is the one being converted to a double.
For: double value2 = 0.999939 * 0.987792; you are multiply two doubles and saving it in a double. This is time there is no casting so the representation never changes (ie. no potential data loss from change in data representation).
For: double value3 = (double)0.999939f * 0.987792f; you are multiplying a double that is casted from a float, times a float. Because of how the Java math works, that second float is probably also being casted to a double, so now you are multiplying two doubles that were once floats causing a third set of representations.
Because float and doubles have different precisions each of these will get a different result. For more info about floating point arithmetic, see here.
When you write a number with the "f" character, it is taken as float, meaning it is encoded with 32bits, whereas without it, it is a double, encoded with 64bits.
The more bits, the more accurately you will represent decimal numbers. This does not make a difference for number with few decimals, but in your case it is significant.
In conclusion, use exclusively double variables in your code.
This question already has answers here:
Why does division result in zero instead of a decimal?
(5 answers)
Closed 8 years ago.
I have hit a snag in my program when this calculates I get result of 0.0
y = 1/6*Math.pow(x,3)+1/2*Math.pow(x,2)-1/3*x;
I have tried writing the equation in chunks so I can add the results up after calculation but the result just keeps ending up being 0.0 and I don't know why. Is this a syntactical error or is there a rule that I'm missing about java?.
When you divide two integers Java truncates the result to an integer. If you want a fractional result you need to use floating point numbers. 1/2 is 0; 1.0/2.0 is 0.5.
y = 1.0/6.0*Math.pow(x,3) + 1.0/2.0*Math.pow(x,2) - 1.0/3.0*x;
y = 1/6*Math.pow(x,3)+1/2*Math.pow(x,2)-1/3*x;
Here you are doing division of two integers, which would result in 0. Make one/both of the values to decimal (1.0/6.0 etc) and then try this. It should give the correct result. The reason is that, 1/6 will be corrected to the closest integer value, which is 0.
This question already has answers here:
Division of integers in Java [duplicate]
(7 answers)
Closed 9 years ago.
I have the following line of code:
System.out.println(5/9);
I expect to see 0.555 as a result, but instead it prints out zero. Can someone help me understand why this happens? I am currently learning programming and appreciate the help.
Thanks!
This happens because what you are unknowingly doing is Integer Division.
To make calculations fast, computer uses Integer division method when there's no decimal number involved, and hence decimal values are lost.
Try this out:
System.out.println(5.0 / 9.0);
or
System.out.println(5.0 / 9);
or
System.out.println(5 / 9.0);
or
System.out.println((float) 5 / 9);
or
System.out.println(5 / (float) 9);
You trying to divide 5 by 9.Both are integers.So the answer is also an integer.So return 0 as answer.So try like System.out.println(5.0/9); or assign values to float variables and go with that
Integer divide by Integer returns Integer
Do like this
System.out.println(5.0/9);
This is an integer division because both operands are of type integer. An integer division gives an integer result obtained by truncation.
When Integer division is ran in Java (Integer / Integer), the number is calculated then rounded down to the while number. Therefore, 5/9 is evaluated as 0.55555 then rounded down to 0.