Why double width = 50/110000; the output is 0.000000000000000? - java

This is my code :
double width = 50/110000;
System.out.println("width ori is "+width );
And the output is: 0.00000000000
What's wrong ? the expected output has to be 4.5454545454545455E-4
Any body can explain to me why?

Because you're dividing two integers, so it will only take the integer part (integer division).
Dividing integers in a computer program requires special care. Some
programming languages, treat integer division (i.e by giving the integer quotient as the answer). So the answer is an integer.
Examples :
In real life In Java
4/3 = 1.33333 4/3 = 1
25/12 = 2.083333 25/12 = 2
9/2 = 4.5 9/2 = 4
50/110000 = 0.000454545 50/110000 = 0
You can cast one of the number (or both but it's actually useless) to double to avoid that :
double width = (double)50/110000;
double width = 50d/110000;
double width = 50.0/110000;

Result of int/int returns you an integer.
So the decimal part got truncated resulting you with an integer
You need to cast:
double width = (double)50/110000;
As #Josh M has pointed, You can also try :
double width = 50d / 110000d;

Explanation to what's happening:
In Java, the default type of numbers is int, so when you write 50/110000, they're both considered int, although you defined the result to be double.
When int division occurs, the result will be 0, because they are both ints, then the double will hold this value, which will be represented as double, so you're getting 0.000000.
Possible solutions:
Coding these numbers with d: 50d/110000d.
Casting one side explicitly (the other will be implicitly cast): (double)50/110000.
50.0/110000.
See Chapter 5. Conversions and Promotions, it'll really help you.

Related

How to get the int value from a floating point number?

If I am getting a random double, how to get only the int?
Examples:
1) 114.999 - get the "114" as an int
2) 565.343234 - get the "565" as an int.
Given the value :
float f = 114.999f;
int i = (int) f;
use a cast to downcast it to an int.
The simplest way would be to cast your double value to an int. For example,
(int) 114.9999 == 114
However, the double type can represent numbers beyond the range of an int. You may need to check if your double is below the smallest possible integer or beyond the largest possible integer to avoid integer overflow issues.
The easiest way to get the integer part of a floating point number would be a simple cast:
double d = 14.999;
int i = (int)d; //14
If you have a primitive wrapper like Double you can use the method intValue() that all subclasses of Number need to provide. However, since those are objects the references can be null and that has to be handled:
Double d = 14.999; //this makes use of auto-boxing
int i = d != null ? d.intValue() : 0; //here 0 is the default value if d is null
Note that this will just truncate the value which can lead to unexpected results due to precision issues, especially when calculations are involved. Due to that you could end up with a number like 14.999999999 when you'd expect 15 or something higher.
Another issue might be that you won't get the next smaller integer for negative values but the next higher, i.e. -14.999 will be truncated to -14.
You should keep that in mind and if those are issues for you have a look at the functions provided by the classes Math, BigDecimal etc.

Java float not acting correctly [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Why does (360 / 24) / 60 = 0 … in Java
I am having this problem:
float rate= (115/100);
When I do:
System.out.println(rate);
It gives me 1.0
What... is the problem?
115 and 100 are both integers, so will return an integer.
Try doing this:
float rate = (115f / 100f);
You're performing integer division (which provides an integer result) and then storing it in a float.
You need to use at least one float in the operation for the result to be the proper type:
float rate = 115f / 100;
float rate= (115/100);
Does the following things:
1) Performs integer division of 115 over 100 this yields the value 1.
2) Cast the result from step 1) to a float. This yields the value 1.0
What you want is this:
float rate = 115.0/100;
Or more generally, you want to convert one of the pieces of your division into a float whether that is via casting (float)115/100 or by appending a decimal point to one of the two pieces or by doing this float rate = 115f / 100 is completely up to you and yields the same result.
In order to perform floating-point arithmetic with integers you need to cast at least one of the operands to a float.
Example:
int a = 115;
int b = 100;
float rate = ((float)a)/b;
use float rate= (float)(115.0/100); instead
It is enough to put float rate = 115f / 100;
The problem you have is that your dividend and divisor are declared as integer type.
In mathematic when you divide two integer results only with remainder. And that is what you assign to your rate variable.
So to have the result as you expected, a remainder with fraction (rational numbers). Your dividend or divisor must be declared in a type with precision.
Base two known types with precision are float (Floating point) and double (Double precision).
By default all numbers (integer literals for purists) written in Java code are in type int (Integer). To change that you need to tell the compiler that a number you want to declare should be represent in different type. To do that you need to append a suffix to integer literal.
Literals for decimal types:
float - f or F; 110f;
double - d or D 110D;
Note that when you would like to use the double, type you can also declare it by adding a decimal separator to literal:
double d = 2.;
or
double d = 2.0;
I encourage you to use double type instead of float. Double type is more suitable for most of modern application. Usage of float may cause unexpected results, because of accuracy problem that in single point calculation have bigger impact on result. Good reading about this “What Every Computer Scientist Should Know About Floating-Point Arithmetic”.
In addition on current CPU architecture both float and double have same performance characteristic. So there is not need to sacrifice the accuracy.
A final note about floating point types in is that non of them should be use when we write a financial application. To have valid results in this matter, you should always used [BigDecimal]

Float to big decimal

I have to write a program with the following requirements:
I have a variable of type float, say float a = 3333.333f;
I have a variable of type int, say int b = 9999;
When I perform a*b in calculator, the result will be 33329996.667
After rounding up the decimals to 2 places, I want to print the value as 33329996.67 in java. I tried with long, double, float, big decimal, But couldnt succeed.
Can anyone please help me solving this?
float only has 7 digits of precision, so its not a good choice for a result with more than 7 digits. double has up to 16 digits of accuracy and is a better choice.
double a = 3333.333;
int b = 9999;
System.out.printf("%.2f", a * b);
prints
33329996.67
To determine the number of digits after the comma you have to apply a little trick:
First shift the comma to the right for the amount of digits you want to have, then cut the whole number e.g. with Math.ceil(float f) and then shift the comma back to the left.
That will illustrate that:
float f = 33329996.667;
float f2 = Math.ceil((f * 100)) / 100;
f2 now has the value 33329996.67.
Hope this helps.
EDIT: For formatting have a look here

different result in java how can it be rectified

Simple calculation gives different result in java.
int a=5363/12*5;
out.println(a);// result is 2230
But actually result should be 2234.5
How can this java result be rectified?
Two issues:
The expression 5363/12*5 gives an integer result (in particular, the division is integer).
The variable a is of type int (integer).
To fix:
double a=5363.0/12*5;
out.println(a);
Note that in general you can't expect to get exact results when using floating-point arithmetic. The following is a very good read: What Every Computer Scientist Should Know About Floating-Point Arithmetic.
5363, 12, and 5 are all being interpreted as ints. the calculation actually being performed here is:
5363/12 = 446.9… - truncated to the int value 446
446 * 5 = 2230
Try specifying a as a float, and indicate that the numbers in the calculation are also created as floats:
float a = 5363f/12f*5f
Take a as double.
Taking a as int will round it to the integer.
Because your all the literal numbers in the right hand side are integers (e.g. 5363 as opposed to 5363.0) expression is being calculated using integer arithmetic semantics i.e. / does whole number division. Thus 5262/12 equals 446 and 446*5 equals 2230. Also your variable a is an int which can only ever hold an integer value.
To fix this you need to do two things. Change the type of a to a decimal type e.g. float or double b) have at least one of 5363 and 12 represented as a decimal type e.g.
double a= 5363.0/12.0*5
Instead of using double you can re-order your expression.
Assuming 5363/12*5 = 5363*5/12 this will give you a closer answer. You have commented you want to round the result so instead you have to add half the value you are dividing by.
int a = (5363 * 5 + /* for rounding */ 6) / 12;
System.out.println(a);
prints
2235
An int is an Integer - nothing after the ..
You should be using
double a = 5363d/12*5;
It seems it has some int/double rounding issue:
double a=((double)5363/12)*5;
System.out.println("VALUE: "+a);
Prints:
VALUE: 2234.5833333333335
Edit: rounding the result to an integer value:
double a=((double)5363/12)*5;
long b=Math.round(a); //you can cast it to an int type if needed
System.out.println("ROUNDED: "+b);
Prints:
ROUNDED: 2235
Use double
double a = 5363/12*5;
System.out.println(a);
or
cast the integer, to prevent loss or precision.
int a = ((int) 5363/12*5);
System.out.println(a);

Error with division using double type in Java

Okay. I have been bashing my head against the wall for like 2 hours now trying to figure out why in the world double answer = 364/365; is telling me that answer is 0. Or any other combination of double for that matter, its just truncating the decimal and I just don't know why.
364/365 performs integer division (truncates the decimal).
Try double answer = 364.0/365; to force it to perform floating point division.
Something like:
double days_in_year = 365;
double answer = 364/days_in_year;
would work as well, since one of the operands isn't an integer.
You're taking an int type (364) and dividing by another int type (365) - the answer is going to be an int. This is then stored in a double type answer. You could do the following:
double answer = 364d / 365d;
More info here:
http://mindprod.com/jgloss/division.html
You need do do double division. Right now Java is interpreting it as integer division and returning the truncated int.
What you need is:
double answer = 364 / 365.0;
or
double answer = 364 / (double) 365;
The reason is that the default type of integer literals in java is int and all the result of all int based arithemetic is type casted back to int. Hence though your answer is 0.997, when it is typecasted back it becomes 0:
(int)0.997 = 0
So you can do like this:
364.0/365.0
or
((float)364)/365
All the above answers are right, would just like to add that it is all about GIGO.
double answer = 364/365;
in above code double type implies only to answer variable and arithmetic expression has both operands of int type. So the output of the arithmetic expression is also int type which is then through auto up-casting to double type gives output 0.0, just like below examples:
double ans = 4.0/0;
the above code will give output Infinity as one of the operand is Floating-point number so through auto type-casting 0 is converted to 0.0 and the result is as per the Floating-point datatype.
whereas
double ans = 4/0;
will give java.lang.ArithmeticException: / by zero exception at runtime since both the operands are of datatype int and so the output is as per the Integer datatype irrespective of ans variable datatype being double.

Categories