Error when using nextInt in Java

Question

I'm working on a program for my classes lab and I seem to be using nextInt wrong and I was wondering if someone could help. I couldn't find a post that had an answer. The program is designed to experiment with loops in a rolling dice simulation. It it supposed to roll two dice, determine the result, and then make a tally if it's snake eyes or doubles or whatever it may be.

So I get the following error:

DiceSimulation.java:33: error: cannot find symbol
     die1Value = nextInt();
                 ^
symbol:   method nextInt()
location: class DiceSimulation

DiceSimulation.java:34: error: cannot find symbol
     die2Value = nextInt();
                 ^
symbol:   method nextInt()
location: class DiceSimulation
2 errors

This is the code I have, I'm not entirely sure how I've used nextInt wrong.

   while (count < NUMBER)
   {
     die1Value = nextInt();
     die2Value = nextInt();
     if (die1Value == die2Value)
     {
        if (die1Value == 1)
        {   
           snakeEyes += 1;
        }
        else if (die1Value == 2)
        {   
           twos += 1;
        }   
        else if (die1Value == 3)
        {   
           threes += 1;
        }   
        else if (die1Value == 4)
        {   
           fours += 1;
        }   
        else if (die1Value == 5)
        {   
           fives += 1;
        }   
        else if (die1Value == 6)
        {   
           sixes += 1;   
        }
      }
      count += 1;
   }

you don't have a method nextInt() in your code, that's what the error is telling you — klor
– klor, Commented Feb 10, 2014 at 1:30
random.nextInt() - you need to specify the appropriate [Random] object instance .. — user2864740
– user2864740, Commented Feb 10, 2014 at 1:30

Makoto · Accepted Answer · 2014-02-10 01:35:32Z

4

Random#nextInt() is neither a static method, nor built-in - you have to have an instance of the Random class to use it.

Here's an example:

Random dice = new Random();
int die1Value = dice.nextInt(6) + 1;
int die2Value = dice.nextInt(6) + 1;

The addition of 1 there is to offset the fact that a random value with a range bound generates values between [0, n).

edited Feb 10, 2014 at 1:35

answered Feb 10, 2014 at 1:29

Makoto

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5 Comments

Vitruvie Over a year ago

Asker wants to roll two dice, not ask for input.

Makoto Over a year ago

So it's not a Scanner, but a Random. Two seconds.

Vitruvie Over a year ago

Still doesn't work; you have to specify the range of the int to return.

MrMike Over a year ago

Indeed, nextInt() is quite ambiguous on its own. I made the same mistake reading the OP's question

Makoto Over a year ago

@Saposhiente: As far as it "working", the signature is correct. As far as it working within the bounds of the program, perhaps you have a point - using nextInt(n) would be correct.

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