Mod(%) operator

[Shiva Mohan]

    
void xxx(){
        for( int x=0;x<4;x++){
            int x1=yyy(x);
            System.out.println("xxxxxx  ------------>>"+x);
        }
    }
    static int yyy(int y){
        int i1=y%2;
        System.out.println("i1i1i1  ------------>>"+i1);
        double d1=y/2;
        System.out.println("d1d1d1 ------------>>"+d1);
       boolean y1=0==y%2;
       System.out.println("yyyyboolean  ------------>>"+y1);
       return y;
    }

When taking y value as 0 in i1,it gives y1 as true.(0==0).When taking y value as 1 in i1,1 mod 2====>1/2=quotient 0.5 and reminder 0.that way I had taken.(calculator return quotient value 0.5.but compiler return quotient value 0.0 and remainder 0.).

I know that mod operator will give remainder part only.But when I take 1%2 working,I always take 0 as the remainder and 0.5 as the quotient.How comw the 1%2 value is quotient 0.0 and remainder 0?

[Rusty Shackleford]

Because they are integer values, not floating point values.

[Campbell Ritchie]

. . . because you are using integer division, which loses its remainder.

     double d1=y/2;

does not produce a double result. It is integer division which produces an integer result and it is then cast to a double.

[Campbell Ritchie]

Posted by Rusty Shackleford

Because they are integer values, not floating point values.

Snap.

[Tony Morris]

Because it's not the mod (or modulus) operator nor does it behave like the modulus operator (which Java does not have).

It's the remainder operator, which is described in the JLS. The rest is explanatory from that point on.