This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 4 years ago.
This is a basic question but I can't find an answer. I've looked into floating point arithmetic and a few other topics but nothing has seemed to address this. I'm sure I just have the wrong terminology.
Basically, I want to take two quantities - completed, and total - and divide them to come up with a percentage (of how much has been completed). The quantities are longs. Here's the setup:
long completed = 25000;
long total = 50000;
System.out.println(completed/total); // Prints 0
I've tried reassigning the result to a double - it prints 0.0. Where am I going wrong?
Incidentally, the next step is to multiply this result by 100, which I assume should be easy once this small hurdle is stepped over.
BTW not homework here just plain old numskull-ness (and maybe too much coding today).
Converting the output is too late; the calculation has already taken place in integer arithmetic. You need to convert the inputs to double:
System.out.println((double)completed/(double)total);
Note that you don't actually need to convert both of the inputs. So long as one of them is double, the other will be implicitly converted. But I prefer to do both, for symmetry.
You don't even need doubles for this. Just multiply by 100 first and then divide. Otherwise the result would be less than 1 and get truncated to zero, as you saw.
edit: or if overflow is likely, if it would overflow (ie the dividend is bigger than 922337203685477581), divide the divisor by 100 first.
In Java
Integer/Integer = Integer
Integer/Double = Double//Either of numerator or denominator must be floating point number
1/10 = 0
1.0/10 = 0.1
1/10.0 = 0.1
Just type cast either of them.
Convert both completed and total to double or at least cast them to double when doing the devision. I.e. cast the varaibles to double not just the result.
Fair warning, there is a floating point precision problem when working with float and double.
If you don't explicitly cast one of the two values to a float before doing the division then an integer division will be used (so that's why you get 0). You just need one of the two operands to be a floating point value, so that the normal division is used (and other integer value is automatically turned into a float).
Just try with
float completed = 50000.0f;
and it will be fine.
As explain by the JLS, integer operation are quite simple.
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).
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.
So to make it short, an operation would always result in a int at the only exception that there is a long value in it.
int = int + int
long = int + long
int = short + short
Note that the priority of the operator is important, so if you have
long = int * int + long
the int * int operation would result in an int, it would be promote into a long during the operation int + long
As your output results a double you should cast either completed variable or total variable or both to double while dividing.
So, the correct implmentation will be:
System.out.println((double)completed/total);
Related
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 4 years ago.
This is a basic question but I can't find an answer. I've looked into floating point arithmetic and a few other topics but nothing has seemed to address this. I'm sure I just have the wrong terminology.
Basically, I want to take two quantities - completed, and total - and divide them to come up with a percentage (of how much has been completed). The quantities are longs. Here's the setup:
long completed = 25000;
long total = 50000;
System.out.println(completed/total); // Prints 0
I've tried reassigning the result to a double - it prints 0.0. Where am I going wrong?
Incidentally, the next step is to multiply this result by 100, which I assume should be easy once this small hurdle is stepped over.
BTW not homework here just plain old numskull-ness (and maybe too much coding today).
Converting the output is too late; the calculation has already taken place in integer arithmetic. You need to convert the inputs to double:
System.out.println((double)completed/(double)total);
Note that you don't actually need to convert both of the inputs. So long as one of them is double, the other will be implicitly converted. But I prefer to do both, for symmetry.
You don't even need doubles for this. Just multiply by 100 first and then divide. Otherwise the result would be less than 1 and get truncated to zero, as you saw.
edit: or if overflow is likely, if it would overflow (ie the dividend is bigger than 922337203685477581), divide the divisor by 100 first.
In Java
Integer/Integer = Integer
Integer/Double = Double//Either of numerator or denominator must be floating point number
1/10 = 0
1.0/10 = 0.1
1/10.0 = 0.1
Just type cast either of them.
Convert both completed and total to double or at least cast them to double when doing the devision. I.e. cast the varaibles to double not just the result.
Fair warning, there is a floating point precision problem when working with float and double.
If you don't explicitly cast one of the two values to a float before doing the division then an integer division will be used (so that's why you get 0). You just need one of the two operands to be a floating point value, so that the normal division is used (and other integer value is automatically turned into a float).
Just try with
float completed = 50000.0f;
and it will be fine.
As explain by the JLS, integer operation are quite simple.
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).
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.
So to make it short, an operation would always result in a int at the only exception that there is a long value in it.
int = int + int
long = int + long
int = short + short
Note that the priority of the operator is important, so if you have
long = int * int + long
the int * int operation would result in an int, it would be promote into a long during the operation int + long
As your output results a double you should cast either completed variable or total variable or both to double while dividing.
So, the correct implmentation will be:
System.out.println((double)completed/total);
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 4 years ago.
This is a basic question but I can't find an answer. I've looked into floating point arithmetic and a few other topics but nothing has seemed to address this. I'm sure I just have the wrong terminology.
Basically, I want to take two quantities - completed, and total - and divide them to come up with a percentage (of how much has been completed). The quantities are longs. Here's the setup:
long completed = 25000;
long total = 50000;
System.out.println(completed/total); // Prints 0
I've tried reassigning the result to a double - it prints 0.0. Where am I going wrong?
Incidentally, the next step is to multiply this result by 100, which I assume should be easy once this small hurdle is stepped over.
BTW not homework here just plain old numskull-ness (and maybe too much coding today).
Converting the output is too late; the calculation has already taken place in integer arithmetic. You need to convert the inputs to double:
System.out.println((double)completed/(double)total);
Note that you don't actually need to convert both of the inputs. So long as one of them is double, the other will be implicitly converted. But I prefer to do both, for symmetry.
You don't even need doubles for this. Just multiply by 100 first and then divide. Otherwise the result would be less than 1 and get truncated to zero, as you saw.
edit: or if overflow is likely, if it would overflow (ie the dividend is bigger than 922337203685477581), divide the divisor by 100 first.
In Java
Integer/Integer = Integer
Integer/Double = Double//Either of numerator or denominator must be floating point number
1/10 = 0
1.0/10 = 0.1
1/10.0 = 0.1
Just type cast either of them.
Convert both completed and total to double or at least cast them to double when doing the devision. I.e. cast the varaibles to double not just the result.
Fair warning, there is a floating point precision problem when working with float and double.
If you don't explicitly cast one of the two values to a float before doing the division then an integer division will be used (so that's why you get 0). You just need one of the two operands to be a floating point value, so that the normal division is used (and other integer value is automatically turned into a float).
Just try with
float completed = 50000.0f;
and it will be fine.
As explain by the JLS, integer operation are quite simple.
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).
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.
So to make it short, an operation would always result in a int at the only exception that there is a long value in it.
int = int + int
long = int + long
int = short + short
Note that the priority of the operator is important, so if you have
long = int * int + long
the int * int operation would result in an int, it would be promote into a long during the operation int + long
As your output results a double you should cast either completed variable or total variable or both to double while dividing.
So, the correct implmentation will be:
System.out.println((double)completed/total);
This question already has answers here:
Int division: Why is the result of 1/3 == 0?
(19 answers)
Closed 4 years ago.
This is a basic question but I can't find an answer. I've looked into floating point arithmetic and a few other topics but nothing has seemed to address this. I'm sure I just have the wrong terminology.
Basically, I want to take two quantities - completed, and total - and divide them to come up with a percentage (of how much has been completed). The quantities are longs. Here's the setup:
long completed = 25000;
long total = 50000;
System.out.println(completed/total); // Prints 0
I've tried reassigning the result to a double - it prints 0.0. Where am I going wrong?
Incidentally, the next step is to multiply this result by 100, which I assume should be easy once this small hurdle is stepped over.
BTW not homework here just plain old numskull-ness (and maybe too much coding today).
Converting the output is too late; the calculation has already taken place in integer arithmetic. You need to convert the inputs to double:
System.out.println((double)completed/(double)total);
Note that you don't actually need to convert both of the inputs. So long as one of them is double, the other will be implicitly converted. But I prefer to do both, for symmetry.
You don't even need doubles for this. Just multiply by 100 first and then divide. Otherwise the result would be less than 1 and get truncated to zero, as you saw.
edit: or if overflow is likely, if it would overflow (ie the dividend is bigger than 922337203685477581), divide the divisor by 100 first.
In Java
Integer/Integer = Integer
Integer/Double = Double//Either of numerator or denominator must be floating point number
1/10 = 0
1.0/10 = 0.1
1/10.0 = 0.1
Just type cast either of them.
Convert both completed and total to double or at least cast them to double when doing the devision. I.e. cast the varaibles to double not just the result.
Fair warning, there is a floating point precision problem when working with float and double.
If you don't explicitly cast one of the two values to a float before doing the division then an integer division will be used (so that's why you get 0). You just need one of the two operands to be a floating point value, so that the normal division is used (and other integer value is automatically turned into a float).
Just try with
float completed = 50000.0f;
and it will be fine.
As explain by the JLS, integer operation are quite simple.
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).
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.
So to make it short, an operation would always result in a int at the only exception that there is a long value in it.
int = int + int
long = int + long
int = short + short
Note that the priority of the operator is important, so if you have
long = int * int + long
the int * int operation would result in an int, it would be promote into a long during the operation int + long
As your output results a double you should cast either completed variable or total variable or both to double while dividing.
So, the correct implmentation will be:
System.out.println((double)completed/total);
I am trying to get exact calculated value (in my case it should be 66.66... not 66.0) but it prints 66.0
If I will get 66.66 then I can use Math.round(66.66) so that I will get 67
Below code after execution should return 66.66 but it returns 66.0
double d = (2*100)/3
Please suggest..
Regards
(2*100)/3 performs integer multiplication and division, which results in an integer.
You need to force floating point calculation by changing one of the operands to an double (or float):
double d = (2.0*100)/3
As other answers suggested , you can provide at least one floating number. Or if you don't want to change the numbers, you can add cast to the numerator
double d= (double)(2*100)/3;
System.out.println(d);
prints
66.66666666666667
As Eran says it's your expression perform integer arithmetic you need to perform like
double d = (double) (2*100)/3;
if needs format the result.
You are using all int values in your arithmetic operation so final result will always be int and when you put it into a double then it becomes 66.0. But since original result is int, so you loose precision.
You can use double d = (2.0*100.0)/3.0; or have at least one value with decimal point, so that you can get expected decimal points.
Number after decimal point is commonly known as precision. So, the issue which you talked about is commonly known as floating-point imprecision, and in your case you can call it as double-point imprecision.
Rule of thumb:
If either is of type double, the other is converted to double for arithmetic operation and final result will be a double.
Otherwise, if either is a float, the other is converted to float for arithmetic operation and final result will be a float.
Otherwise, if either is of type long, the other is converted to long for arithmetic operation and final result will be a long.
Otherwise, both operands are converted to int for arithmetic operation and final result will be a int.
I am new to Java, and wondering how does double to int cast work ? I understand that it's simple for long to int by taking the low 32 bits, but what about double (64 bits) to int (32 bits) ? those 64 bits from double in binary is in Double-precision floating-point format (Mantissa), so how does it convert to int internally ?
It's all documented in section 5.1.3 of the JLS.
In the first step, the floating-point number is converted either to a long, if T is long, or to an int, if T is byte, short, char, or int, as follows:
If the floating-point number is NaN (§4.2.3), the result of the first step of the conversion is an int or long 0.
Otherwise, if the floating-point number is not an infinity, the floating-point value is rounded to an integer value V, rounding toward zero using IEEE 754 round-toward-zero mode (§4.2.3). Then there are two cases:
If T is long, and this integer value can be represented as a long, then the result of the first step is the long value V.
Otherwise, if this integer value can be represented as an int, then the result of the first step is the int value V.
Otherwise, one of the following two cases must be true:
The value must be too small (a negative value of large magnitude or negative infinity), and the result of the first step is the smallest representable value of type int or long.
The value must be too large (a positive value of large magnitude or positive infinity), and the result of the first step is the largest representable value of type int or long.
(The second step here is irrelevant, when T is int.)
In most cases I'd expect this to be implemented using hardware support - converting floating point numbers to integers is something which is usually handled by CPUs.
Java truncates its value if you use (int) cast, as you may notice:
double d = 2.4d;
int i = (int) d;
System.out.println(i);
d = 2.6;
i = (int) d;
System.out.println(i);
Output:
2
2
Unless you use Math.round, Math.ceil, Math.floor...
You may want to read the Java specification:
http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html#jls-5.1.3
The relevant section is 5.1.3 - "Narrowing Primitive Conversion".