Java rounding up to an int using Math ceil

Question

int total    int  Math ceil 157 32     Why does it still return 4  157 32   4 90625  I need to round up  I ve looked around and this seems to be the right method   I tried total as double type  but get 4 0   What am I doing wrong

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int total    157-1  32   1   or more general   a-1  b  1

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Java provides only floor division   by default  But we can write ceiling in terms of floor  Let s see   Any integer y can be written with the form y    q k r  According to the definition of floor division  here floor  which rounds off r   floor q k r  k     q    where 0   r   k-1   and of ceiling division  here ceil  which rounds up r1   ceil q k r1  k     q 1    where 1   r1   k   where we can substitute r 1 for r1   ceil q k r 1  k     q 1    where 0   r   k-1     Then we substitute the first equation into the third for q getting  ceil q k r 1  k     floor q k r  k    1    where 0   r   k-1   Finally  given any integer y where y   q k r 1 for some q k r  we have  ceil y  k     floor y-1  k    1   And we are done  Hope this helps

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Also to convert a number from integer to real number you can add a dot   int total    int  Math ceil 157 32      And the result of  157 32   will be real too

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Check the solution below for your question   int total    int  Math ceil 157 32     Here you should multiply Numerator with 1 0  then it will give your answer   int total    int  Math ceil 157 1 0 32

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There are two methods by which you can round up your double value    Math ceil  Math floor   If you want your answer 4 90625 as 4 then you should use Math floor and if you want your answer 4 90625 as 5 then you can use Math ceil  You can refer following code for that   public class TestClass        public static void main String   args            int floorValue    int  Math floor  double 157   32           int ceilValue    int  Math ceil  double 157   32           System out println  Floor    floorValue           System out println  Ceil    ceilValue

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When dividing two integers  e g    int c    int  a    int  b   the result is an int  the value of which is a divided by b  rounded toward zero  Because the result is already rounded  ceil   doesn t do anything  Note that this rounding is not the same as floor    which rounds towards negative infinity  So  3 2 equals 1  and floor 1 5  equals 1 0  but  -3  2 equals -1  but floor -1 5  equals -2 0    This is significant because if a b were always the same as floor a    double  b   then you could just implement ceil   of a b as -   -a    b    The suggestion of getting ceil a b  from  int n    a   b - 1    b   which is equivalent to a   b    b - 1    b  or  a - 1    b   1  works because ceil a b  is always one greater than floor a b   except when a b is a whole number  So  you want to bump it to  or past  the next whole number  unless a b is a whole number  Adding 1 - 1   b will do this  For whole numbers  it won t quite push them up to the next whole number  For everything else  it will   Yikes  Hopefully that makes sense  I m sure there s a more mathematically elegant way to explain it

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int total    int  Math ceil   double 157   double  32

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Use double to cast like   Math ceil  double value  or like  Math ceil  double value1  double value2

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157 32 is int int  which results in an int   Try using the double literal - 157 32d  which is int double  which results in a double

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You are doing 157 32 which is dividing two integers with each other  which always result in a rounded down integer  Therefore the  int  Math ceil      isn t doing anything  There are three possible solutions to achieve what you want  I recommend using either option 1 or option 2  Please do NOT use option 0  Option 0 Convert a and b to a double  and you can use the division and Math ceil as you wanted it to work  However I strongly discourage the use of this approach  because double division can be imprecise  To read more about imprecision of doubles see this question  int n    int  Math ceil  double  a   b     Option 1 int n   a   b     a   b    0    0   1     You do a   b with always floor if a and b are both integers  Then you have an inline if-statement witch checks whether or not you should ceil instead of floor  So  1 or  0  if there is a remainder with the division you need  1  a   b    0 checks for the remainder  Option 2 This option is very short  but maybe for some less intuitive  I think this less intuitive approach would be faster than the double division and comparison approach  Please note that this doesn t work for b  lt  0  int n    a   b - 1    b   To reduce the chance of overflow you could use the following  However please note that it doesn t work for a   0 and b  lt  1  int n    a - 1    b   1   Explanation behind the  quot less intuitive approach quot  Since dividing two integer in Java  and most other programming languages  will always floor the result  So  int a  b  int result   a b  is the same as floor a b     But we don t want floor a b   but ceil a b   and using the definitions and plots from Wikipedia   With these plots of the floor and ceil function you can see the relationship    You can see that floor x   lt   ceil x   We need floor x   s    ceil x   So we need to find s  If we take 1 2  lt   s  lt  1 it will be just right  try some numbers and you will see it does  I find it hard myself to prove this   And 1 2  lt    b-1    b  lt  1  so ceil a b    floor a b   s              floor a b    b-1  b              floor   a b-1  b     This is not a real proof  but I hope your are satisfied with it  If someone can explain it better I would appreciate it too  Maybe ask it on MathOverflow

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Nobody has mentioned the most intuitive   int x    int  Math round Math ceil  double  157   32      This solution fixes the double division imprecision

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int total    int  Math ceil  double 157 32

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157 32 is an integer division because all numerical literals are integers unless otherwise specified with a suffix  d for double l for long   the division is rounded down  to 4  before it is converted to a double  4 0  which is then rounded up  to 4 0   if you use a variables you can avoid that  double a1 157  double a2 32  int total    int  Math ceil a1 a2

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I know this is an old question but in my opinion  we have a better approach which is using BigDecimal to avoid precision loss  By the way  using this solution we have the possibility to use several rounding and scale strategies  final var dividend   BigDecimal valueOf 157   final var divisor   BigDecimal valueOf 32   final var result   dividend divide divisor  RoundingMode CEILING  intValue

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In Java adding a  0 will make it a double     int total    int  Math ceil 157 0   32 0

[java] Java rounding up to an int using Math.ceil

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