Division of integers in Java

Question

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

User · Answer

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);

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

User · Answer

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

[java] Division of integers in Java

Examples related to java

Examples related to math

Examples related to floating-point

Examples related to long-integer

Examples related to println