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Why does the division of two integers return 0.0 in Java? [duplicate]
(6 answers)
Why does this Java division print out zero? [duplicate]
(5 answers)
Closed 8 years ago.
I am using Java 1.6
final double check = 3 / 4;
System.out.println(check);
Console is showing: 0.0
Why is this happening? Shouldn't it come out 0.75?
Make that:
double check = 3.0 / 4;
and it'll work. You got 0 because 3 / 4 is an integer division, whose value is 0.
Because both are integer hence result will also be integer.
Cast any one into double like this:
double check = (double)3 / 4;
By doing:
3 / 4
you are performing an integer division, since 3 and 4 are int constants, not doubles. The result is therefore an int. But since you are assigning it to a double, the result of the division will then be promoted to a double. But at this point it is too late, since the integer division will have produced 0!
You need to do:
3.0 / 4
to achieve your desired result, since in this case, 4 will automatically be promoted to a double, and the result of the division will also be a double.
To be perfectly sure of what happens and if you like symmetry, you can also write:
3.0 / 4.0
You are dividing integers and assigning the result to double.In java division of two int values always yields an int.
So change the statement double check = 3 / 4; to double check = 3.0 / 4;
Related
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:
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.
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Dividing by 100 returns 0
(6 answers)
Closed 9 years ago.
Would someone mind explaining why this doesn't work? All variables except for chance are ints, whereas chance is a double. When I print all the values they are definitely correct... but chance always comes out as 0.0. I know this has something to do with converting ints to doubles, as I have had an issue like this a couple of times before. What is the key to getting it to do what you want?
gladValue = (glad.dexterity+glad.tactical+weaponSkill);
oppValue = (glad.target.dexterity+glad.target.tactical+glad.target.agility);
chance = (gladValue/oppValue)*10.0;
Thanks
You should write gladValue * 10.0 / oppValue instead.
The reason is quite subtle. Your brackets mean that gladValue / oppValue is computed first. But these variables are integers so the result is an integer and therefore you lose the fraction part. Only when it is multipled by 10.0 will it get promoted to a double; but by then it's too late.
If you do as I say then, bearing mind that * and / have the same precedence and the operations happen from left to right, then when computing gladValue * 10.0, gladValue is promoted to floating point and that floating point result is divided by oppValue.
This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Why does the division of two integers return 0.0 in Java?
I have some very confusing problem..
I want to calculate some stuff and after some debugging I saw that java calculates this arithmetic: 2 / 4 = 0.0 But it should be 0.5
2 & 4 are stored in integer variables
the result is stored in a double-type.
Did I miss something clearly?
It is because of integer division (Integer division rounds toward 0). Cast one of the operand to double type.
Example:
double temp = (double)2/4
will give you correct results.
Use
myDouble = (double) integerWhoseValueIs2 / integerWhoseValueIs4
The fact you store it in a double doesn't change the fact that the division of two integers makes an integer. When you store it, it's too late.
From the Java Language Specification :
Integer division rounds toward 0. That is, the quotient produced for
operands n and d that are integers after binary numeric promotion
(§5.6.2) is an integer value q whose magnitude is as large as possible
while satisfying |d · q| ≤ |n|.
If you do not write a cast to double, the result is always an int.
double a = (double) 2/4;