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?
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.