Java program error: possible loss of precision

问题:

I am trying to figure out why I am getting this loss of precision error in my Java program.

This is the error:

error: possible loss of precision
         int digit = num/Math.pow(10,i-1)%10;
                                         ^
  required: int
  found:    double
1 error

This is the code clip inside a for loop

for(int i = 1; i<=String.valueOf(num).length(); i++)
{
    int digit = num/Math.pow(10,i-1)%10;
    //etc
}

I want to keep digit as an integer variable, and I tried to declare digit outside the loop, and I also tried casting it into an integer which didn't work. I did try doing it as:

digit += num/Math.pow(10,i-1)%10; //with digit declare outside of loop

which worked magically, but I don't want to increment it, instead I want it to contain the digit which I can use for later. Is there something I'm missing here? Please explain.

回答1:

I tried to run your code as follows:

int digit = (int) (num/Math.pow(10,i-1)%10);

And it worked as expected, i.e. it iterated through the digits of your variable num (which I assumed as int).

When you use the += operator, it works not because of magic, but because of automatic type promotion. You can read more about it here: http://docs.oracle.com/javase/specs/jls/se7/html/jls-5.html

回答2:

The easist fix is to simply cast

num/Math.pow(10,i-1)%10; 

into an Integer. You can do this by adding (int) before it:

 int digit = (int) num/Math.pow(10,i-1)%10;

As to why this happens: Math.pow is defined as a double and won't convert a Double value into an Integer value without explicitly telling it to.

回答3:

Math.pow returns a double, and all your other operands in the expression are integral types. To fix the compiler error, explicitly cast the result of Math.pow to an int, as indicated by the other answers.

Side note. Don't use Math.pow to find 10 to the power of i-1, because floating-point exponentiation is unnecessarily expensive. Instead, multiply by 10 on every iteration, like so:

int r = 1;
for(int i = 1; r < num; i++, r *= 10) {
    int digit = num / r % 10;
    // ...
}