Why Does Math.pow(x,y) Count as a Double?

Question

I'm writing a Java program to calculate how much food it will take to get a monster to a certain level in My Singing Monsters. When I run the program, it says, "cannot convert from double to int". Can someone explain why this is? Here's the program.

int totalFood = 0;
int level = 1;
int levelMeal = 5*(Math.pow(2,level-1));
int mealNumber = 1;
int levelGoal = 1;
while(level != levelGoal)
{
  if(mealNumber != 5)
  {
    mealNumber += 1;
    totalFood += levelMeal;
  }
  else if(mealNumber == 5)
  {
    mealNumber = 0;
    level += 1;
  }
}
if(level == levelGoal)
{
  println("The total amount of food required for a monster to reach level " + levelGoal + " is " + totalFood + " food.");
}

[Here's](http://docs.oracle.com/javase/7/docs/api/java/lang/Math.html#pow(double,double)) the javadoc. — Sotirios Delimanolis, Apr 27 '14 at 17:54
When the error is displayed, the line that says "int levelMeal = 5*(Math.pow(2,level-1));" is highlighted. — Jason Chen, Apr 27 '14 at 17:54
@JasonChen What datatype pow returns? And to what type of value are you trying to assign it? When you'll answer those questions, the answer will be obvious. — Alexis C., Apr 27 '14 at 17:55

score 18 · Accepted Answer · answered Apr 27 '14 at 17:57

18

You'll have to do this:

int levelMeal = (int) (5*(Math.pow(2,level-1)));
                  ^
           this is a cast

As you can see in the documentation, Math.pow() returns a double by design, but if you need an int then an explicit cast must be performed.

answered Apr 27 '14 at 17:57

Óscar López

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`5<<(level-1)` would be a more appropriate implementation (if you fix the same overflow problems you get from going through double), and will be more precise for very high `level`s (if you'd use `long` or even a bignum type instead of `int`). – Apr 27 '14 at 18:03
@Rhymoid - That's what I put in my answer. However, I overlooked the fact that I could just directly shift the 5 instead of doing a multiply afterwards. I'll update my answer. – DaoWen Apr 27 '14 at 18:04

DaoWen · Answer 2 · 2014-04-27T18:16:47.027

4

I think there's typically hardware support on most modern processors for doing floating-point powers, but not integers. Because of that, for a general power, it's actually faster to do Math.power with a double and then convert it back to an int.

However, in this case there's a faster way to do it for ints. Since you're doing a power of 2, you can just use the bitwise left-shift operator instead:

int levelMeal = 5*(1<<(level-1));

As Rhymoid pointed out in his comment, that expression can be further simplified to remove the 1:

int levelMeal = 5<<(level-1);

edited Apr 27 '14 at 18:16

answered Apr 27 '14 at 18:01

DaoWen

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Just `5<<(level-1)` will do too. I doubt javac is clever enough to make that optimization by itself. – Apr 27 '14 at 18:05

score 0 · Answer 3 · edited Apr 27 '14 at 18:00

0

Math.pow return double and you assigning double value to int this is why it is giving error. You have to downcast it. Like

int levelMeal = (int)5*(Math.pow(2,level-1));

edited Apr 27 '14 at 18:00

keyser

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answered Apr 27 '14 at 17:59

niiraj874u

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Why Does Math.pow(x,y) Count as a Double?

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