This question already has answers here:
Why does the division of two integers return 0.0 in Java? [duplicate]
(6 answers)
Double value returns 0 [duplicate]
(3 answers)
Closed 9 years ago.
Why does Java return a 0 when I divide 10/60?
the code I tried is
double pay_per_minute = (10/60);
10 being the pay rate per hour and 60 being the minutes.
Because you're building an integer. When you store it in a double variable, it's already too late : it's 0.
Do
double pay_per_minute = (10.0/60);
If you have variables, cast them :
double pay_per_minute = ((double)pay_per_hour) / 60;
any primitive digit in java is treated as an integer, so when you do 10/60, it is integer division and due to precision loss it is giving 0
Here 10 and 60 takes as int values and then you get int dividing result it is 0 then you get answer as 0. use following way.
double a=10;
double b=60;
double div=a/b;
you need to type cast it first because by default numericals are considered as integers
double pay_per_minute = ((double)10/60);
System.out.println(pay_per_minute);
output 0.16666666666666666
double pay_per_minute = (10/60);
Here, you are dividing integer 10 by integer 60. So, this is like doing
int temp = 10/60;
double pay_per_minute = double(temp)
your temp will be 0 (since 10/60 is 0 when considered as integer division)
You need to do,
double pay_per_minute = (10.0/60);
Related
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 3 years ago.
when i want to get the result of 100/100000..
i got only 0.
Example
int one = 100;
int two = 100000;
int result = one/two;
toast(result); //Result is 0
Hey there "int" data type only stores integer values and not the decimals.
So if you divide 3 with 2 you would get 1 as answer instead of 1.5 .
Int just ignores the decimals .
You need to choose float or double data type for this to work.
Your variable named result must be declared and casted to float data type.
Appreciate the effort and mark this as answer if it helps you.....
This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Division of integers in Java [duplicate]
(7 answers)
Closed 5 years ago.
Why is it that in the first line sevenTwelfths will evaluate to the expected answer (0.5833), but threeTwentySixths will evaluate to zero? I assumed that since the data type is a double, the operation of dividing 3 by 26 would be a decimal, but it appears as if it does the operation as an integer operation and then coverts that answer to a double and stores it in threeTwentySixths.
double sevenTwelfths = ((double) 7 / 12);
double threeTwentySixths = 3 / 26;
double sevenTwelfths = ((double) 7 / 12);
In First line this called typecasting meaning you are converting your answer in double for example
int sevenTwelfths=((int)7/12));
this mean after dividing 7/12 whatever is the anser convert into int or Type cast into int
This question already has answers here:
Always Round UP a Double
(8 answers)
Closed 7 years ago.
Is there a way to automatically round up a value in java?
For example:
//generate a random integer value
int randomVal = RandomHelper.nextIntFromTo(1, otherVal);
/* then divide the integer value in half... the error I am getting is that its a double, probably
because the number generated isn't an even number, but I NEED it to be an integer. Can I round up?*/
int value = randomVal * 0.5;
You can add 1 then divide by 2 instead of multiplying by 0.5. That way, you avoid a floating point operation followed by a conversion to int.
int value = (randomVal + 1) / 2;
Use Math.ceil():
int value = (int)Math.ceil(randomVal * 0.5);
This question already has answers here:
How to round up the result of integer division?
(18 answers)
How to round up integer division and have int result in Java? [duplicate]
(9 answers)
Closed 4 years ago.
I know I can use Math.java functions to get the floor, ceil or round value for a double or float, but my question is -- Is it possible to always get the higher integer value if a decimal point comes in my value
For example
int chunkSize = 91 / 8 ;
which will be equal to 11.375
If I apply floor, ceil or round to this number it will return 11 I want 12.
Simply If I have 11.xxx I need to get 12 , if I have 50.xxx I want 51
Sorry The chunkSize should be int
How can I achieve this?
Math.ceil() will do the work.
But, your assumption that 91 / 8 is 11.375 is wrong.
In java, integer division returns the integer value of the division (11 in your case).
In order to get the float value, you need to cast (or add .0) to one of the arguments:
float chunkSize = 91 / 8.0 ;
Math.ceil(chunkSize); // will return 12!
ceil is supposed to do just that. I quote:
Returns the smallest (closest to negative infinity) double value that
is greater than or equal to the argument and is equal to a
mathematical integer.
EDIT: (thanks to Peter Lawrey): taking another look at your code you have another problem. You store the result of integer division in a float variable: float chunkSize = 91 / 8 ; java looks at the arguments of the division and as they are both integers it performs integer division thus the result is again an integer (the result of the division rounded down). Now even if you assign this to the float chunkSize it will still be an integer(missing the double part) and ceil, round and floor will all return the same value. To avoid that add a .0 to 91: float chunkSize = 91.0 / 8;. This makes one of the arguments double precision and thus double division will be performed, returning a double result.
If you want to do integer division rounding up you can do
x / y rounded up, assuming x and y are positive
long div = (x + y - 1) / y;
In your case
(91 + 8 - 1) / 8 = 98 / 8 = 12
This works because you want to round up (add one) for every value except an exact multiple (which is why you add y - 1)
Firstly, when you write float chunkSize = 91 / 8 ; it will print 11.0 instead of 11.375
because both 91 and 8 are int type values. For the result to be decimal value, either dividend or divisor
or both have to be decimal value type. So, float chunkSize = 91.0 / 8; will result 11.375 and
now you can apply Math.ceil() to get the upper value. Math.ceil(chunkSize); will result in 12.0
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 9 years ago.
So, I'm doing a fighting game in java and I have this equation that always returns zero.
int x = 90/100*300;
It should be 270 but it returns zero. :|
You're doing integer calculation, so 90/100 results in 0.
If you write it 90.0/100*300, the calculations will be done with doubles (then you'll need to cast it back to int if you want).
The problem here is it first divides 90/100 which is actually 0.9 however since it is an int type the value is 0 and then when you multiply 0 with 300 the output is 0.
90/100 is 0 remainder 90.
You can fix this by re-arranging the values
int x = 90 * 300 / 100; // 270
Do the multiplications before the divisions and as long as you don't get an overflow, this will be more accurate.
Java in this case carries out the multiplication/division in order.
It also interpretes each number as integers.
So, 90/100 = 0.9 and it gets truncated to 0.
and 0 * 300 = 0.
So you end up with 0
You use integer datatype you use float or double datatype and get proper answer of this equation. Integer datatype only decimal value answer return but float and double datatype return answer with floating point then must use float or double datatype in this equation.
I guess you want like this equation, try this
int x = (90/100)*300;