Related
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);
}
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);
}
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);
}
I have values like this:
long millis = 11400000;
int consta = 86400000;
double res = millis/consta;
The question is: why res equals 0.0 (instead of ca. 0.131944)? It's stored in double so there should be no rounding right?
When you are using a binary operator, both arguments should be of a same type and the result will be in their type too. When you want to divide (int)/(long) it turns into (long)/(long) and the result is (long). you shouldmake it (double)/(long) or (int)/(double) to get a double result. Since double is greater that int and long, int and long will be turned into double in (double)/(long) and (int)/(double)
Because you are dividing a long by an int you get an long results.
What you are effectively doing is
double res = (double) (millis/consta);
as millis/consta is 0, when cast to double is 0.0
Try the following to divide a double by an int and get a double result.
double res = (double) millis/consta;
which is the same as
double res = ((double) millis)/((double) consta));
You are doing longdivision (int gets cast to long) so you get long values, which are integers (so, 0)
You should do
double res = (double) millis / consta;
Once one of the values is casted to double, the other is too casted so the operation uses the same type in both operators.
millis/consta is an integer division, which results in 0. the casting in the line:
double res = millis/consta;
is done on the result:
double res = (double)(millis/consta);
What you need to do is to cast one of the operands:
double res = (double)millis/consta;
The resulting type of a long and int devision will be a long, which can't hold decimals.
you want to cast it to a double before you assign it
When this happens it automatically downcasts to int to perform the operation. Thy writing it like:
long millis = 11400000;
int consta = 86400000;
double res = ((double)millis)/((double)consta);
And it will work.
This is because
long: The long data type is a 64-bit signed two's complement integer.
int: The int data type is a 32-bit signed two's complement integer.
See Primitive type
It means both are integer.
Since we divide long by integer, result is long 0.
But you assign this value to res(double) value and print it.
So the result 0.0 is shown.
long millis = 11400000;
int consta = 86400000;
System.out.println("millis/consta = " + millis/consta); // print 0
double res = millis/consta;
System.out.println("Res " + res); // print 0.0
In Java, I want to convert a double to an integer, I know if you do this:
double x = 1.5;
int y = (int)x;
you get y=1. If you do this:
int y = (int)Math.round(x);
You'll likely get 2. However, I am wondering: since double representations of integers sometimes look like 1.9999999998 or something, is there a possibility that casting a double created via Math.round() will still result in a truncated down number, rather than the rounded number we are looking for (i.e.: 1 instead of 2 in the code as represented) ?
(and yes, I do mean it as such: Is there any value for x, where y will show a result that is a truncated rather than a rounded representation of x?)
If so: Is there a better way to make a double into a rounded int without running the risk of truncation?
Figured something: Math.round(x) returns a long, not a double. Hence: it is impossible for Math.round() to return a number looking like 3.9999998. Therefore, int(Math.round()) will never need to truncate anything and will always work.
is there a possibility that casting a double created via Math.round() will still result in a truncated down number
No, round() will always round your double to the correct value, and then, it will be cast to an long which will truncate any decimal places. But after rounding, there will not be any fractional parts remaining.
Here are the docs from Math.round(double):
Returns the closest long to the argument. The result is rounded to an integer by adding 1/2, taking the floor of the result, and casting the result to type long. In other words, the result is equal to the value of the expression:
(long)Math.floor(a + 0.5d)
For the datatype Double to int, you can use the following:
Double double = 5.00;
int integer = double.intValue();
Double perValue = 96.57;
int roundVal= (int) Math.round(perValue);
Solved my purpose.