This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Closed 7 years ago.
When doing the following in Java, i get as the result 0.0. Even though the result should be 10. This is an example calculation, in my code it's doing this with with double values.
double result = (10 / 100) * 100;
10 and 100 are integer literals, so 10 / 100 is evaluated in integer arithmetic which means that any remainder is discarded.
So you get (10 / 100) * 100 = (0) * 100
One fix is to promote one of the arguments to a double:
(10.0 / 100) * 100
Try this:
double result = (10.0 / 100) * 100
You are working with integer and int 10 / int 100 is 0. 0 * 100 is also 0
Related
I found this formula on SO but I don't get how it works:
(n + 5) / 10 * 10
This rounds a number to the multiple of ten but when I calculate it it just divides and multiplies back to n+5. How does this actually work?
Example (in Java):
(24 + 5) / 10 * 10 = 20
While it should actually be 29 as far as I know.
You're doing these operations on int. That's why, every operation result is also an int. What happens is:
(24 + 5) = 29
29 / 10 = 2 (because the fraction part is always dropped in ints)
2 * 10 = 20
This rounding method works fine only for float and double. You can try to cast the int to double.
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:
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);
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 4 years ago.
Is there any way to calculate (for example) 50% of 120?
I tried:
int k = (int)(120 / 100)*50;
But it doesn't work.
int k = (int)(120 / 100)*50;
The above does not work because you are performing an
integer division expression (120 / 100) which result is
integer 1, and then multiplying that result to 50, giving
the final result of 50.
If you want to calculate 50% of 120, use:
int k = (int)(120*(50.0f/100.0f));
more generally:
int k = (int)(value*(percentage/100.0f));
int k = (int)(120*50.0/100.0);
Never use floating point primitive types if you want exact numbers and consistent results, instead use BigDecimal.
The problem with your code is that result of (120/100) is 1, since 120/100=1.2 in reality, but as per java, int/int is always an int.
To solve your question for now, cast either value to a float or double and cast result back to int.
I suggest using BigDecimal, rather than float or double. Division by 100 is always exact in BigDecimal, but can cause rounding error in float or double.
That means that, for example, using BigDecimal 50% of x plus 30% of x plus 20% of x will always sum to exactly x.
it is simple as 2 * 2 = 4 :)
int k = (int)(50 * 120) / 100;
Division must be float, not int
(120f * 50 / 100f)
You don't need floating point in this case you can write
int i = 120 * 50 / 100;
or
int i = 120 / 2;