CODEWITHSUNDEEP
javapow
peter
peter

Reputation: 8473

Java pow implementation with Integer.MIN_VALUE as exponent

I am implementing a pow function in Java, and I am wondering how do we deal with Integer.MIN_VALUE as a exponent ? Do we just treat it as a special case ?

Because I tried to compare the result with the standard Java.lang.Math API and I get a couple different result. The following is the list of comparison

//this will print "1.0 vs 0.0"
System.out.println(pow(2,Integer.MIN_VALUE) + " vs " + Math.pow(2,Integer.MIN_VALUE));

//this will print "1.0 vs 1.0"
System.out.println(pow(1,Integer.MIN_VALUE) + " vs " + Math.pow(1,Integer.MIN_VALUE));

public double pow(double base, int exp){
     double result = 1.0;
     boolean pos = false;

     if(exp == 0) return result;
     if(exp > 0){
         pos = true;
         exp *= -1;
     }

     while(exp > 0){
        if((exp & 1) == 1){
           result *= base;
        }
        base *= base;
        exp /= 2;
     }
     if(!pos){
        result = 1/result;
     }
     return result;
}

So I am wondering if Integer.MIN_VALUE is a special case where I have to have a if statement for checking it.

  if(exp == Integer.MIN_VALUE && base > 1) return 0.0;

Upvotes: 2

Views: 650

Answers (4)

Partha Mishra
Partha Mishra

Reputation: 312

Well, the real issue is that, since the sign doesn't flip on the MIN_VALUE, the sign cascades to the exp/2. and the 'negative power' case applies. If we split it, it's easier:

    public double myPow(double x, int n) {
    double result = 1.00000;
    boolean negative = false;
    
    if(n <0) {
        negative = true;
        n= -n;
    }
    
    result=power(x,n);
    if(negative) {
        result = 1/result;
    }
    return result;
}

private double power(double a, int n) {
    if(n ==0 || a==1) return 1;// a^0 = 1, 1^n = 1
    double x=power(a,n/2);
    
    if(n%2 == 0) return x*x;
    else return a*x*x;
}

Upvotes: 0

djechlin
djechlin

Reputation: 60788

On my system I have

-2147483648
2147483647

For Integer.MIN_VALUE and Integer.MAX_VALUE respectively. So you should see the problem in the line

exp *= -1;

Upvotes: 0

Louis Wasserman
Louis Wasserman

Reputation: 198221

Yeah, you've got the problem that Integer.MIN_VALUE * -1 == Integer.MIN_VALUE. You could either special-case it, or you could deal with it another way. Indeed, one possible solution would be to make exp negative when it's positive, instead of the other way around; you'd just use -exp instead of exp.

Upvotes: 0

Joe K
Joe K

Reputation: 18434

Based on this line:

exp *= -1;

it seems that it might have to be a special case. There are certainly ways to implement this without that special case, but because -1 * Integer.MIN_VALUE cannot be stored in an int, you will get a bug if you do not handle it separately.

Upvotes: 2

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