I was writing this code:
public static void main(String[] args) {
double g = 1 / 3;
System.out.printf("%.2f", g);
}
The result is 0. Why is this, and how do I solve this problem?
The two operands (1 and 3) are integers, therefore integer arithmetic (division here) is used. Declaring the result variable as double just causes an implicit conversion to occur after division.
Integer division of course returns the true result of division rounded towards zero. The result of 0.333... is thus rounded down to 0 here. (Note that the processor doesn't actually do any rounding, but you can think of it that way still.)
Also, note that if both operands (numbers) are given as floats; 3.0 and 1.0, or even just the first, then floating-point arithmetic is used, giving you 0.333....
1/3 uses integer division as both sides are integers.
You need at least one of them to be float or double.
If you are entering the values in the source code like your question, you can do 1.0/3 ; the 1.0 is a double.
If you get the values from elsewhere you can use (double) to turn the int into a double.
int x = ...;
int y = ...;
double value = ((double) x) / y;
Explicitly cast it as a double
double g = 1.0/3.0
This happens because Java uses the integer division operation for 1 and 3 since you entered them as integer constants.
Because you are doing integer division.
As #Noldorin says, if both operators are integers, then integer division is used.
The result 0.33333333 can't be represented as an integer, therefore only the integer part (0) is assigned to the result.
If any of the operators is a double / float, then floating point arithmetic will take place. But you'll have the same problem if you do that:
int n = 1.0 / 3.0;
The easiest solution is to just do this
double g = (double) 1 / 3;
What this does, since you didn't enter 1.0 / 3.0, is let you manually convert it to data type double since Java assumed it was Integer division, and it would do it even if it meant narrowing the conversion. This is what is called a cast operator.
Here we cast only one operand, and this is enough to avoid integer division (rounding towards zero)
The result is 0. Why is this, and how do I solve this problem?
TL;DR
You can solve it by doing:
double g = 1.0/3.0;
or
double g = 1.0/3;
or
double g = 1/3.0;
or
double g = (double) 1 / 3;
The last of these options is required when you are using variables e.g. int a = 1, b = 3; double g = (double) a / b;.
A more completed answer
double g = 1 / 3;
This result in 0 because
first the dividend < divisor;
both variables are of type int therefore resulting in int (5.6.2. JLS) which naturally cannot represent the a floating point value such as 0.333333...
"Integer division rounds toward 0." 15.17.2 JLS
Why double g = 1.0/3.0; and double g = ((double) 1) / 3; work?
From Chapter 5. Conversions and Promotions one can read:
One conversion context is the operand of a numeric operator such as +
or *. The conversion process for such operands is called numeric
promotion. Promotion is special in that, in the case of binary
operators, the conversion chosen for one operand may depend in part on
the type of the other operand expression.
and 5.6.2. Binary Numeric Promotion
When an operator applies binary numeric promotion to a pair of
operands, each of which must denote a value that is convertible to a
numeric type, the following rules apply, in order:
If any operand is of a reference type, it is subjected to unboxing
conversion (§5.1.8).
Widening primitive conversion (§5.1.2) is applied to convert either or
both operands as specified by the following rules:
If either operand is of type double, the other is converted to double.
Otherwise, if either operand is of type float, the other is converted
to float.
Otherwise, if either operand is of type long, the other is converted
to long.
Otherwise, both operands are converted to type int.
you should use
double g=1.0/3;
or
double g=1/3.0;
Integer division returns integer.
Make the 1 a float and float division will be used
public static void main(String d[]){
double g=1f/3;
System.out.printf("%.2f",g);
}
The conversion in JAVA is quite simple but need some understanding. As explain in the JLS for integer operations:
If an integer operator other than a shift operator has at least one operand of type long, then the operation is carried out using 64-bit precision, and the result of the numerical operator is of type long. If the other operand is not long, it is first widened (§5.1.5) to type long by numeric promotion (§5.6).
And an example is always the best way to translate the JLS ;)
int + long -> long
int(1) + long(2) + int(3) -> long(1+2) + long(3)
Otherwise, the operation is carried out using 32-bit precision, and the result of the numerical operator is of type int. If either operand is not an int, it is first widened to type int by numeric promotion.
short + int -> int + int -> int
A small example using Eclipse to show that even an addition of two shorts will not be that easy :
short s = 1;
s = s + s; <- Compiling error
//possible loss of precision
// required: short
// found: int
This will required a casting with a possible loss of precision.
The same is true for the floating point operators
If at least one of the operands to a numerical operator is of type double, then the operation is carried out using 64-bit floating-point arithmetic, and the result of the numerical operator is a value of type double. If the other operand is not a double, it is first widened (§5.1.5) to type double by numeric promotion (§5.6).
So the promotion is done on the float into double.
And the mix of both integer and floating value result in floating values as said
If at least one of the operands to a binary operator is of floating-point type, then the operation is a floating-point operation, even if the other is integral.
This is true for binary operators but not for "Assignment Operators" like +=
A simple working example is enough to prove this
int i = 1;
i += 1.5f;
The reason is that there is an implicit cast done here, this will be execute like
i = (int) i + 1.5f
i = (int) 2.5f
i = 2
1 and 3 are integer contants and so Java does an integer division which's result is 0. If you want to write double constants you have to write 1.0 and 3.0.
I did this.
double g = 1.0/3.0;
System.out.printf("%gf", g);
Use .0 while doing double calculations or else Java will assume you are using Integers. If a Calculation uses any amount of double values, then the output will be a double value. If the are all Integers, then the output will be an Integer.
Because it treats 1 and 3 as integers, therefore rounding the result down to 0, so that it is an integer.
To get the result you are looking for, explicitly tell java that the numbers are doubles like so:
double g = 1.0/3.0;
(1/3) means Integer division, thats why you can not get decimal value from this division. To solve this problem use:
public static void main(String[] args) {
double g = 1.0 / 3;
System.out.printf("%.2f", g);
}
public static void main(String[] args) {
double g = 1 / 3;
System.out.printf("%.2f", g);
}
Since both 1 and 3 are ints the result not rounded but it's truncated. So you ignore fractions and take only wholes.
To avoid this have at least one of your numbers 1 or 3 as a decimal form 1.0 and/or 3.0.
My code was:
System.out.println("enter weight: ");
int weight = myObj.nextInt();
System.out.println("enter height: ");
int height = myObj.nextInt();
double BMI = weight / (height *height)
System.out.println("BMI is: " + BMI);
If user enters weight(Numerator) = 5, and height (Denominator) = 7,
BMI is 0 where Denominator > Numerator & it returns interger (5/7 = 0.71 ) so result is 0 ( without decimal values )
Solution :
Option 1:
doubleouble BMI = (double) weight / ((double)height * (double)height);
Option 2:
double BMI = (double) weight / (height * height);
I noticed that this is somehow not mentioned in the many replies, but you can also do 1.0 * 1 / 3 to get floating point division. This is more useful when you have variables that you can't just add .0 after it, e.g.
import java.io.*;
public class Main {
public static void main(String[] args) {
int x = 10;
int y = 15;
System.out.println(1.0 * x / y);
}
}
Do "double g=1.0/3.0;" instead.
Many others have failed to point out the real issue:
An operation on only integers casts the result of the operation to an integer.
This necessarily means that floating point results, that could be displayed as an integer, will be truncated (lop off the decimal part).
What is casting (typecasting / type conversion) you ask?
It varies on the implementation of the language, but Wikipedia has a fairly comprehensive view, and it does talk about coercion as well, which is a pivotal piece of information in answering your question.
http://en.wikipedia.org/wiki/Type_conversion
Try this out:
public static void main(String[] args) {
double a = 1.0;
double b = 3.0;
double g = a / b;
System.out.printf(""+ g);
}
Using the formula "(4/3)*3.14*pow(radius,3)", the number returned always yields to an integer. Any ideas on how to get a floating point number from the formula?
Code:
#include<iostream>
#include<math.h>
using namespace std;
int main(){
float radius, v;
cout<<"Enter value of radius";cin>>radius;
v = (4/3)*3.14*pow(radius, 3);
cout<<"volume of sphere: "<<v<<endl;
system("pause");return 0;
}
The answer is 1 because, 4 and 3 are integers, and in C++ integer divided by integer yields another integer, so 4/3 = 1.
But to correct that, all you need to do is this,
Change the integer to a floating point decimal number, for e.g. 4 could be written as 4.0 instead.
(4.0/3)*3.14*pow(radius,3)
Thats all and you are good to go.
The expression 4/3 performs integer division, and will result in 1 (it results in truncation of any decimals). You should be performing floating point division.
v = (4.0/3.0)*3.14*pow(radius, 3);
As 4 and 3 both are integers as their division will also be an integer, to obtain the floating point value as a result, you need to specify both the values to be float (i.e 4.0 and 3.0)
By applying above you will end up something like this
(4.0/3.0)=floating point value.
This gives output as 0:
print -4/-5
Whereas:
print float(-4/-5)
This gives output as 0.0 . The required output is 0.8
You are doing integer division instead of floating point division. It has been answered already: Python division .
Casting types after the division doesn't make sense.
float(4)/float(5)
Or simpler
4./5.
should do the trick
To understand,
print float(-4/-5)
Bracket is calculated first. Value given to float is 0. Typecasting 0 to 0.0
This will give the required output:
print float(-4)/-5
/ does integer division.
To get your desired output, the operands should be float (either or both).
-4.0 / -5.0 = 0.8
To explain the second code snippet, the first one to be evaluated is the operation -4 / -5 which results to 0 since we did an integer division. Now what you tried to do is to convert 0 to a floating point using the function float(). Converting that resulted to 0.0
I'm trying to calculate the 8th root square of a value or its ^1/8, but numpy is always returning the wrong value
temp = 141.18
h2 = temp ** (1/8)
h2_ = np.power(temp, (1/8))
my output is always 1.0 .
I've tried square command too.
I need to use numpy, I'm using other numpy arrays in mycode, just to keep compatible.
>>> 1/8
0
>>> 1./8
0.125
And of course, anything to the power of 0 results in 1.
Understand the numeric tower.
Rule 1: Given two operands of the same type, the result will have that type.
e.g. int / int = int
temp**(1/8) does not give the 8th root of temp because:
>>>1/8
0
Rule 2: If the operands are mixed, one of them will be coerced up the numeric tower: integer --> rational --> float --> complex.
e.g. float / int = float
>>>1./8 # 1. is a float
0.125
Note: There may be cases where these rules do not apply to true division / and floor division // but I don't fully understand them. See the link.
"They've done studies you know. It works 60% of the time... everytime." - Brian Fantana
Trap: In the OPs question the expression temp**(1/8) is made of mixed operands (temp is a float) so why isn't (1/8) a float?
The operands are evaluated according to BODMAS/BIDMAS so (1/8) is evaluated first, the resulting expression becomes temp**0 and at this point 0 is coerced to a float.
Any positive int or float to the power 0.0 is 1.0.
In python, i cannot divide 5 by 22. When I try this, it gives me zero-even when i use float?!!
>>> print float(5/22)
0.0
It's a problem with order of operations. What's happening is this:
* First python takes 5/22. Since 5 and 22 are integers, it returns an integer result, rounding down. The result is 0
* Next you're converting to a float. So float(0) results in 0.0
What you want to do is force one (or both) operands to floats before dividing. e.g.
print 5.0/22 (if you know the numbers absolutely)
print float(x)/22 (if you need to work with a variable integer x)
Right now you're casting the result of integer division (5/22) to float. 5/22 in integer division is 0, so you'll be getting 0 from that. You need to call float(5)/22.