Java优先级

Question

int x = 1;
System.out.println( x++ + x++ * --x );

The above code prints out "5" but I don't understand how? I keep getting zero for the last x which is then multiplied by the second x which is still 0 and then I get 2? Please help!

Answer 1

The way the program processes your statement is as follows:

x = 1;
1 + (increment x) 2 * (increment x)(decrement x) 2 =
1 + 2 * 2 =
1 + 4     =
5

Added:

If you ask it to print it out for you instead of actually doing the arithmetic, you'll see what the x's actually equal:

    int x = 1;
    System.out.println(x++ + " + " +  x++  + "*" + --x);

Output: 1 + 2*2

Answer 2

It works like this:

System.out.println( x++ + x++ * --x );

Since the first two are postfix they will not be performed until after a value is already put in. A 1 is put in the first x then the value is increased to 2. A 2 is put in the second x and the value is increased to 3.

System.out.println( 1 + 2 * --x );

Since the --x is prefix the operation is done prior to subbing in the value. Therefore it would equal 2 and x would equal 2 again.

System.out.println( 1 + 2 * 2 );

After this it works the same as it normally would in math (multiplication before addition).

Answer 3

public static void main(String[] args) {

        int x ;
        x = 1; 

        System.out.println( x++ );
        //1

        System.out.println( x++ * --x );
        //2*--3 = 2*2 = 4

        System.out.println( x++ + (x++ * --x ));
        //1 + 4 = 5

       System.out.println( x++ + x++ * --x );
       //1 + 4 = 5
    }