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Closed 10 years ago.
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Why does int exp1 = 14/20*100; equals ‘0’ in java?
I am using this:
int newWidth = Math.round(297 * (40 / 414));
In the code the numbers are actually variables, but those are what they were holding, and when I run the code newWidth comes back as 0, and I was expecting 28 or 29. I don't see what is wrong here...
40 / 414 is rounded to 0 immediately as it only works with integers. You must either cast one of the operands to double/float or use double immediately:
int newWidth = (int)Math.round(297 * (40.0 / 414));
or:
int newWidth = (int)Math.round(297 * ((double)40 / 414));
Because 40/414 equals to 0, so 297*0 = 0 and Math.round(0) = 0
Use doubles instead.
int newWidth = (int)Math.round(297 * (40d / 414));
Change to int newWidth = (int)Math.round(297 * (40.0 / 414));
An int in java uses no decimal places... Therefore you should use a double for this. It is giving you a 0 because inside the () 40/ 414 = 0.
The problem starts with 40/414: that's integer division, which returns an int. In this case: 0.
To fix it, cast one of those ints to a double or float, e.g. ((float) 40 / 414).
The result of 40/414 is zero. This is a consequence of integer division. Change either numerator or denominator to a floating point value to get your desired result
Gonna throw my 2 cents in... as noted, the issue lies in the integer division. Another method for forcing this division to treat the result as a double:
int newWidth = Math.round(297 * (40.0 / 414));
40/414 is rounded to 0.
What you probably want is
int newWidth = (297 * 40) / 414;
Related
I can't seem to find the answer I'm looking for regarding a simple question: how do I round up any number to the nearest int?
For example, whenever the number is 0.2, 0.7, 0.2222, 0.4324, 0.99999 I would want the outcome to be 1.
So far I have
int b = (int) Math.ceil(a / 100);
It doesn't seem to be doing the job, though.
Math.ceil() is the correct function to call. I'm guessing a is an int, which would make a / 100 perform integer arithmetic. Try Math.ceil(a / 100.0) instead.
int a = 142;
System.out.println(a / 100);
System.out.println(Math.ceil(a / 100));
System.out.println(a / 100.0);
System.out.println(Math.ceil(a / 100.0));
System.out.println((int) Math.ceil(a / 100.0));
Outputs:
1
1.0
1.42
2.0
2
See http://ideone.com/yhT0l
I don't know why you are dividing by 100 but here my assumption int a;
int b = (int) Math.ceil( ((double)a) / 100);
or
int b = (int) Math.ceil( a / 100.0);
int RoundedUp = (int) Math.ceil(RandomReal);
This seemed to do the perfect job. Worked everytime.
10 years later but that problem still caught me.
So this is the answer to those that are too late as me.
This does not work
int b = (int) Math.ceil(a / 100);
Cause the result a / 100 turns out to be an integer and it's rounded so Math.ceil
can't do anything about it.
You have to avoid the rounded operation with this
int b = (int) Math.ceil((float) a / 100);
Now it works.
Just another option. Use basics of math:
Math.ceil(p / K) is same as ((p-1) // K) + 1
The easiest way to do this is just:
You will receive a float or double and want it to convert it to the closest round up then just do System.out.println((int)Math.ceil(yourfloat));
it'll work perfectly
Assuming a as double and we need a rounded number with no decimal place . Use Math.round() function.
This goes as my solution .
double a = 0.99999;
int rounded_a = (int)Math.round(a);
System.out.println("a:"+rounded_a );
Output :
a:1
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 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;
I'm trying to round the users input but I can seem to get my double to round to an int. Basically, when I enter 4.4999 it wont round up to 5.
Any ideas?
Math.ceil() returns the ceiled value. It can't change the value of the variable it takes as argument, because Java passes arguments by value. So you need to do
hours = Math.ceil(hours);
The actual solution is to use double inside the ceil method.
Math.ceil(7 * 50 / 100) will return 3.0 even though the actual value resulting from 7*50/100 is 3.5. It is because since everything is int, the result of 350/100 itself will be 3.
If however, if you give Math.ceil(7 * 50 / 100D), the result would be 4.0.
So, the 4.999 in your question should be a double and not a result of an integer operation like 4999/1000.
Just make sure that whatever you give inside a the ceil is double and not an int.
Both function return the rounded (or ceiled) values, but don't change the variable passed as parameter.
Use eg. hours = Math.ceil(hours);.
Math.ciel returns a Double. Something like this should work (inside of your hours > 0 block):
cost += Math.ceil(hours) * hourlyRate;
You're not assigning the result of Math.ceil(hours) back to hours so it will never round.
int a = 15
int b = 2;
int x = (int) Math.ceil( a / b );
int y = (int) Math.ceil( (double) a / (double) b );
Results:
x: 7
y: 8
I'm trying to do some basic math and it keeps popping up as 0. I'm sure it has to do with it being an int but I don't know how to work around it. I need to use integers but the math to arrive at those integers uses decimals. How do I do it?
That's integer division.
To get non-integer results, use doubles instead.
This is not special to blackberry, it's standard java behaviour.
This is because you're doing integer math:
int subexpr1 = 14 / 20; // 0
int subexpr2 = subexpr1 * 100; // 0
Use a double instead or change the order
int expr1 = (int) 14.0/20 * 100; // Very small possibility of rounding errors
int expr2 = 14 * 100 / 20; // Will ignore fraction parts
You can change it to 14*100/20 - and then it will give what you want.
I.e. change the sequence of operations (14/20 is 0)
Your result is being cast as an int, so you are losing precision.
Try
double exp1 = 14/20.0*100;