||遇到麻烦 和&&

Question

I am new to programming and wanted to make a dice rolling programm in Java for execise. The code is the following:

import java.math.*;
public class Dices {
public static int dice1=0;
    public static int dice2=0;
    public static int x;
    public static void main(String args[]){
    do {
        x++;
        dice1=(int) (Math.random()*6+1);
        dice2=(int) (Math.random()*6+1);
        System.out.println(dice1+", "+dice2);
    } while(dice1 !=1 || dice2 !=1);
    System.out.println("Finalthrow: "+dice1+", "+dice2);
    System.out.println("Snake-Eyes after "+x+" tries.");
    }
}

This way it works fine, but in my opinion there is something wrong with the code. In the while condition should actually be. But if I use && it stops as soon as it rolls a 1 on the first dice. I thought && means "AND" and || means "OR". So actually it should behave exactly the other way around, or am I misinterpreting something?

Answer 1

Some understanding about Morgan's laws could help here. The law says (sorry for the weird syntax, but I think the message is clear) that :

(!P) OR (!Q) == !(P AND Q)
(!P) AND (!Q) == !(P OR Q)

So when you use || (OR) in your condition

while(dice1 !=1 || dice2 !=1)

is exactly the same as

while(!(dice1 == 1 && dice2 == 1))

so it will looop until both dice are 1 .

On the other hand, if you use && (AND) :

while(dice1 !=1 && dice2 !=1)

it's the same as

while(!(dice1 == 1 || dice2 == 1))

so it means that it will loop until one or two of the dice is/are 1 .

Answer 2

&& means and
|| means or

So (dice1 != 1 || dice2 != 1) means continue the loop while dice1 is not 1 or dice 2 is not 1.

So (dice1 != 1 && dice2 != 1) means continue the loop while dice1 is not 1 and dice 2 is not 1.

Answer 3

The code is fine. You want the loop to end when dice1 == 1 and dice2 == 1 . So it must loop until that is true, or until its opposite is false. The opposite of dice1 == 1 && dice2 == 1 is !(dice1 == 1 && dice2 == 1) which is equivalent to dice1 != 1 || dice2 != 1 dice1 != 1 || dice2 != 1 .

Think of it this way: if dice1 != 1 , keep looping. Also, if dice2 != 1 , keep looping. So if either is true, keep looping. And to test if either is true, regardless of if both are true, use || .

Answer 4

The behavior is wright. You're saying with dice1 !=1 && dice2 !=1 that repeat the loopuntil BOTH of the dices are NOT 1. But when one dice rolls a 1 the condition is false and the loop escapes. Try it with a truth table.

Answer 5

Let's make a truth table.

What can we conclude from this? If you want your loop to continue while either A or B are different from 1 (or, as De Morgan's laws say: until both of them are equal to 1 ) then you can narrow it down to these values:

Since we need to find the operator that allows us to continue the loop (aka: the condition is true ), we take the column that returns true for all 3 different kinds of inputs, which is A || B A || B .

Note that A && B and A || B A || B refer to the result of A != 1 and B != 1 , not the actual input.