Int division Why is the result of 1 3 0

Question

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

User · Answer

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

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you should use  double g 1 0 3    or  double g 1 3 0    Integer division returns integer

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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

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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

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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

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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

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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

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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

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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

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Do  double g 1 0 3 0   instead

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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

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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  lt  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     quot Integer division rounds toward 0  quot  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

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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

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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 - gt  long int 1    long 2    int 3  - gt  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 - gt  int   int - gt  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   lt - 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

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I did this  double g   1 0 3 0  System out printf  quot  gf quot   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

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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

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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

[java] Int division: Why is the result of 1/3 == 0?

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