This question already has answers here:
How to make the division of 2 ints produce a float instead of another int?
(9 answers)
Closed 9 years ago.
I'm trying to do a very basic operation as follows:
double a=21/5;
System.out.println(a);
However, each time I get 4.0 as the output and not 4.2. I'm encountering this for the first time. I've been using Java for years, but never came across this obscurity.
You are using integer division, which result will always be integer
You should use something like this.
double a=(double)21/5;
You are doing integer division...
Try:
double a = 21.0/5;
Cast the division or specify one of the arguments as a decimal to force the return as a double:
double a = (double)21/5;
-or-
double a = 21.0/5;
Just cast one of the numbers to double:
double a = 21/5.0;
Force the cast to double.
double a = 21.0/5
This is called Arithmetic promotion. This means that all terms in an equation are made equal to the variable type with the highest precision. In this case double.
Related
This question already has answers here:
Division of integers in Java [duplicate]
(7 answers)
Closed 4 years ago.
I want the help to produce the result of below calculation as 1.12 but the result is coming up 1.0
double k=(112)/100;
System.out.println(k);
You are doing Integer division causing it to lose the precision:
Replace
double k=(112)/100;
with
double k=(112.0)/100;
double k-=((double)112/100) worked for me as suggested by agni
when we do division like (112/100) JVM gives int as an o/p, you are storing that 'int' in 'double' so JVM adds '.0' to the o/p causing loss of precision. so, here you have to tell the JVM to give o/p without any loss. for that reason we have to mention like `double k = (double)112/100; which is similar to "typecasting".
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
How to make the division of 2 ints produce a float instead of another int?
(9 answers)
Division of integers in Java [duplicate]
(7 answers)
Closed 5 years ago.
So I'm trying to do math on ints stored in an array of ints.
float day1Hours = (day1[3]-day1[2]) / 2;
for this specific problem, day1[3] = 19 and day1[2] is 10. So, it should be doing 19-10 = 9, and then dividing that by 2 to make 4.5. But, the output that I am getting is 4. I've also tried storing day1Hours as a double but that made no difference. How would I make this be able to do the math correctly and get those decimal values that I need?
The problem is that you are doing integer division and then converting to a float. Try
float day1Hours = (day1[3]-day1[2]) / 2.0f;
Using a float literal in the denominator will cause the division to be done in floating point, and you won't get integer truncation. (As an alternative to using a float literal, you could cast the numerator or denominator to a float, but that seems somewhat baroque to me. It would be more suitable if both the numerator and denominator were int variables.)
The reason that just changing the type of day1Hours doesn't affect the problem is that the entire right side is evaluated first using the declared data type of day1 and then converted to whatever type is on the left of the assignment.
float day1Hours = (float)(20-9) / 2; //5.5
Problem is that on the right side of equation the numbers are integers and when dividing 2 integers the decimal places are truncated (integer division rounds towards zero) 4.5 -> 4.0.
Try changing 2 -> 2f so the 2 would be considered a float instead of an integer.
This question already has answers here:
How to make the division of 2 ints produce a float instead of another int?
(9 answers)
Closed 7 years ago.
double multiply()
{
double x=(2/3)*3.14*1.02;
System.out.print(x);
double y=0.666*3.14*1.02; /*(2/3)=0.666...*/
System.out.print(y);
}
Output:
x=0.0
y=SomeNumber
please explain this?
(2/3) is 0.
because both are integers. To solve this, use a cast or make it clear that your number is not an integer:
double x=(2/3d)*3.14*1.02;
Now you have an integer divided by a double, which results in a double.
Some more to read about this:
http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/04/mixed.html
(2/3) is computed first (because of the parentheses), and in integer arithmetic (since the number literals are of type int). The fractional part is discarded.
It is therefore an int type with a value of 0. The entire expression is therefore zero.
The obvious remedy is to remove the parentheses and write 2.0 / 3.0 instead. Some folk prefer an explicit cast, but I find that ugly.
2/3 = 0 because they don't have explicit cast to double they are integers. The whole expression becomes: double x=0*3.14*1.02;
which is 0.
because data type of both 2 and 3 is int and int/int gives you int which in your case 2/3 is 0.
Try using 2.0/3 or 2/3.0 you will get the required answer.
Suffix every number with a 'd' to be sure you are dealing with doubles
This question already has answers here:
Java program error: possible loss of precision
(3 answers)
Closed 6 years ago.
int Asum,temp4;
temp4=Math.sqrt(Asum);
This is part of a code I'm writing. I've declared Asum and temp4 as integers. I'm trying to get the sqrt of Asum and assign it to temp4. but for some reason, java gives me the error : Possible loss of precision. reqd=int. found= double.
I need temp4 and Asum to be an integer. and it's not necessary to get the precise decimal values, I just need the rounded off integers.
Have you tried to add an explicit cast to int so that the compiler knows that you know what you are doing?
temp4= (int) Math.sqrt(Asum);
temp4 shouldn't be integer... it could be float.
I think you need to make use of the Math.round function.
Try:
temp4=Math.round(Math.sqrt(Asum));
This question already has answers here:
Double division behaving wrongly
(4 answers)
Closed 7 years ago.
Noob Question.
double answer = 13/5;
System.out.println(answer);
Why does that return 2 instead of 2.6.
As in must i say
(double) 13/5
everytime i want to print a double.
you must tell java that you want a double division:
double answer = 13/5.0; // or 13 / (double) 5.0
13 and 5 are integers. That means that you divide them as integers. As a result, you will get an integer. So what you have to do is to cast them as double like that:
double answer = 13 / (double) 5;
Or write them like that:
double answer = 13 / 5.0;
Because 13 and 5 are integers, so the result will be an integer that you are assigning to a double variable
You need to specify your numbers should be considered as double
The division takes place between two ints and the result is also an int which is finally cast into a double. You should declare the operands as doubles to get the desired result.