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Java casting (cast operator expressions) [duplicate]

This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Closed 6 years ago.
The question require me to write a Java program to show the results of the following cast operator expressions:
(double) (23 / 14) + 7.65
My Code:
public class op {
public static void main(String [] args) {
int num = 23/14;
double r1 = (double) num;
double result = r1 + 7.65;
System.out.println("Results: "+ result);
}
}
I don't think I have done correctly, what are the problems of my code?
By the way, can someone tell me what are the differences between long, double, int, float? How do we know when to use these primitive data types? I read an explanation here: https://docs.oracle.com/javase/tutorial/java/nutsandbolts/datatypes.html
but is there any 'human-version' of the explanation?
Thank you for your help.

The problem is due to the used types.
Since you divide two integers (23 and 14), the result is considered and int as well. Therefor, 23/14 = 1.642857142857143, which is truncated to fit in an int result, more specifically, 1.
result is the sum of 1 (int) and 7.65 (double). Since one of them is a double, to other is converted to the "upper" type as well (double) and the operation becomes 1.0+7.65 = 8.65.
The result is correct, because you asked the result of (double) (23 / 14) + 7.65 which means the result of casting the result of the operations in brackets to double summed with 7.65. Which is 8.65 as previously explained.
If you want to use a division using doubles, consider:
double r1 = 1.0 * 23/14;

Lets see step-by-step:
int num = 23/14; // int division of 23/14 results in 1
So, here num = 1
When you cast num to double value of r1 is setted to 1.0.
double result = r1 + 7.65; //1.0 + 7.65 = 8.65

ok, int is short term for INTEGER which are natural numbers that we use normally but with no decimal places and if your number has some value in between roughly -2 billion to +2 billion. if your range exceeds that and you still want an integer then go for long data type.
floats are for decimal values like 3.147 with a range of +10*38 to -10*38 or so, but if your range exceeds this(practically this happens rarely) go for double.
coming to the code you put here , if you divide a int by another int (like 23/14) you get only get the integer part of the answer(only '1' in 23/14=1.642...) , next when you cast it to double you get 1.0 and next you are going to add that to 7.65 which will make the ultimate answer as 8.65 hope this answers your Q....

You could change this int num = 23/14
to double num = ((double) 23)/14
or double num = (23 * 1.0)/14

Java integer-double division confusion [duplicate]

This question already has answers here:
Integer division: How do you produce a double?
(11 answers)
Closed 7 years ago.
Program 1
int sum = 30;
double avg = sum / 4; // result is 7.0, not 7.5 !!!
VS.
Program 2
int sum= 30
double avg =sum/4.0 // Prints lns 7.5
Is this because the '4' in program 1 is acting as a literal integer? so 30/4 would give me 7. However since this data type is a double, we need to add a .0 to end. so '7.0'
Program 2 has 4.0 which is acting as a literal double. an int/double would always give double because it more precise. so we get '7.5'. I don't understand what double data type is doing to the result though.. it really doesn't need to do anything since the conditions of the double data type are already satisfied.(have the most precise result out of the computation).
Am I wrong? I encourage you to correct me! This is how I learn.. :)

In your first example:
int sum = 30;
double avg = sum / 4; // result is 7.0, not 7.5 !!!
sum is an int, and 4 is also an int. Java is dividing one integer by another and getting an integer result. This all happens before it assigns the value to double avg, and by then you've already lost all information to the right of the decimal point.
Try some casting.
int sum = 30;
double avg = (double) sum / 4; // result is 7.5
OR
int sum = 30;
double avg = sum / 4.0d; // result is 7.5

This is an integer division (because it involves two integers)
int sum = 30;
double avg = (sum / 4); // result is 7
Integer division, will round down to the nearest integer.
However, this is a double division (because 4.0 is a double)
int sum= 30
double avg = (sum/4.0) // result is 7.5
This is the expected behaviour, and the conversion are all well defined. No need to guess. Take a look at the java docs about conversion.
A widening conversion of an int or a long value to float, or of a long
value to double, may result in loss of precision - that is, the result
may lose some of the least significant bits of the value. In this
case, the resulting floating-point value will be a correctly rounded
version of the integer value, using IEEE 754 round-to-nearest mode

In java, operations involving only integer types (int, long and so on) have integer results. This includes divisions.
Both 30 and 4 are integer values, so if you try to divide them, integer division is used.
But if you try either 30.0/4 or 30/4.0 then the result will be 7.5 because you will be using floating-point division.
The declared type of the variable avg has no influence on the result. The decimal part is lost during the division, and not when you assign the value to the variable.

Image taken from : http://www.mathcs.emory.edu/~cheung/Courses/170/Syllabus/04/mixed.html
Refer above URL for more clear explanation.

PROGRAM 1
In Java, when you do a division between two integers, the result is an integer. So when you do sum/4, the result is the integer 7. Then you assign that integer to a double variable, so that 7 turns into 7.0.
PROGRAM 2
In Java, when you do a division between an integer and a double, the result is a double. So when you do sum/4.0, the result is the double 7.5. Then you assign that to a double variable, so it's still 7.5.
So the problem is not how println prints the variables, but how division works in Java

how divided integer is converted to floating point number with decimal [duplicate]

This question already has answers here:
Division of integers in Java [duplicate]
(7 answers)
Closed 7 years ago.
if 123/33 prints out 3 and 3 is an integer if we cast it to float ( (float)123/33 )how do we get the decimal places from the integer 3. does 3 contains floating points internally what ?
public static void test() {
System.out.println("==========");
System.out.println(123/33); // prints 3
System.out.println((float)123/33); // prints 3.7272727
//so if we cast it to float we get the decimal points also (.7272727)
}

The cast does not apply to the entire expression 123/33. You're casting the value 123 to a float, which causes any further operations to use floating-point math.

123 will be casted(converted) to float and then the division happens, so the result will be float value.
System.out.println("==========");
System.out.println(123/33); // prints 3
System.out.println((float)123); // prints 123.0
System.out.println((float)123/33); // prints 3.7272727

Dividing error in java language

I have a simple java program that does not operate the way that I think it should.
public class Divisor
{
public static void main(String[] args)
{
int answer = 5 / 2;
System.out.println(answer);
}
}
Why is this not printing out 2.5?

5 / 2 is integer division (you're even storing it in an integer variable), if you want it to be 2.5, you need to use floating point division:
double answer = 5.0 / 2.0;
Integer division is always going to be equal to normal mathematical division rounded down to the nearest integer.

Java has integer division which says: integer divided by integer results in integer. 2.5 cannot be represented with integer so the result is floored to 2.0. Moreover, you store the result in integer.
If you need floating point division you can cast one of operands to double and change answer type to double as well. You use literal values here, so changing 5 to 5. makes this literal value double.
In the end the following should work for you:
double answer = 5. / 2;
Note, you don't even need a zero sign after a dot symbol!

Floating point in Java [duplicate]

This question already has answers here:
Closed 10 years ago.
Possible Duplicate:
Why does (360 / 24) / 60 = 0 … in Java
Having problems with floats and longs in Java
float f = 0.100f;
f = 3/180;
At the minute i am trying to do something like this with object and their attributes, but even to this simplest form my program returns 0.0.
I have tried this with Longs as well as still the same result. It's been a long day and maybe it's something simple but I'm at a brick wall.

Your expression 3/180 is performing an integer division, and it is then casting that into the float f. In integer division, 3/180 will return 0, and this is what you are seeing.
What you probably want to do is just add a decimal point to your numbers: f = 3.0/180.0;

3/180 is integer division.
Therefore, the result is truncated to an integer.
You need to perform floating-point division: 3/180f

You do an integer division. So you get an int which is casted back to a float. Try this:
f = 3/180.0;
or
f = 3/180f;

Try 3.0/180. Otherwise, you are dividing two integers and you run into integer truncation. When you do integer division the result is also an integer, not a floating point number.