问题
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?
回答1:
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....
回答2:
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;
回答3:
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.
回答4:
you should use
double g=1.0/3;
or
double g=1/3.0;
Integer division returns integer.
回答5:
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;
回答6:
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;
回答7:
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);
}
回答8:
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
回答9:
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.
回答10:
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.
回答11:
(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);
}
回答12:
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.
回答13:
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);
}
回答14:
Do "double g=1.0/3.0;" instead.
回答15:
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
来源：https://stackoverflow.com/questions/4685450/int-division-why-is-the-result-of-1-3-0