为什么我的析因程序打印为0？

Question

This code is meant to find 100 factorial and the sum the digits in the number. However, it returns 0.

Why does this happen?

public class Problem20 {
  public static void main(String[] args){
    int num = 100;

    for(int i = 1; i< 100; i++){
        num = num * i;
    }

    String numstring = Integer.toString(num);
    int sum = 0;

    for(int j = 0; j < numstring.length(); j++){
        sum += numstring.charAt(j) - '0';
    }
    System.out.print(sum);
  }
}

Answer 1

Every time you multiply by 2, you add a 0 to the low bits of binary representation of the number.

According to the JLS §15.17.1 :

If an integer multiplication overflows, then the result is the low-order bits of the mathematical product as represented in some sufficiently large two's-complement format. As a result, if overflow occurs, then the sign of the result may not be the same as the sign of the mathematical product of the two operand values.

Here's some code to demonstrate this point. As you can see, the number of 0 at the end of the number slowly increases; soon as we get to 32, the low order bits become all 0. See: Why does this multiplication integer overflow result in zero?

public class Problem20 {
  public static void main(String[] args) {
    int num = 100;

    for (int i = 1; i < 33; i++) {
      num = num * i;
      System.out.println(i + "\t" + Integer.toBinaryString(num));
    }
  }
}

Output:

1   1100100
2   11001000
3   1001011000
4   100101100000
5   10111011100000
6   10001100101000000
7   1111011000011000000
8   1111011000011000000000
9   10001010011011011000000000
10  10101101000010001110000000000
11  11101101111011000011010000000000
12  100111000100100111000000000000
13  11111011111011111011000000000000
14  11000111000110111010000000000000
15  10101010100111100110000000000000
16  10101001111001100000000000000000
17  1001000010001100000000000000000
18  10100111011000000000000000000
19  10001101100001000000000000000000
20  1110010100000000000000000000
21  101100100100000000000000000000
22  11010100011000000000000000000000
23  10100101000000000000000000000
24  11101111000000000000000000000000
25  1010111000000000000000000000000
26  11010110000000000000000000000000
27  10010010000000000000000000000000
28  11111000000000000000000000000000
29  11000000000000000000000000000
30  11010000000000000000000000000000
31  110000000000000000000000000000
32  0

You can solve this problem by using BigInteger instead of int , eg

import java.math.BigInteger;

public class Problem20 {
  public static void main(String[] args) {
    BigInteger num = BigInteger.valueOf(100);

    for (int i = 1; i < 100; i++) {
      num = num.multiply(BigInteger.valueOf(i));
      System.out.println(num);
    }
  }
}

Answer 2

Alright.

1! = 1
2! = 2
3! = 6
4! = 24
.
.
.
10! = 3628800
.
.
100! = 93326215443944152681699238856266700490715968264381621468592963895217599993229915608941463976156518286253697920827223758251185210916864000000000000000000000000

Do you think this fits in any of those data-types?

What happens once a number crosses the limit? It overflows. And hence you get a wrong answer.

So what do I do?

Use BigInteger .