为什么Java认为从10到99的所有数字的乘积都是0?

2022-08-31 08:23:00

以下代码块将输出为 0。

public class HelloWorld{

    public static void main(String []args){
        int product = 1;
        for (int i = 10; i <= 99; i++) {
            product *= i;
        }
        System.out.println(product);
    }
}

请有人解释为什么会发生这种情况?


答案 1

以下是程序在每个步骤中执行的操作:

          1 * 10 =          10
         10 * 11 =         110
        110 * 12 =        1320
       1320 * 13 =       17160
      17160 * 14 =      240240
     240240 * 15 =     3603600
    3603600 * 16 =    57657600
   57657600 * 17 =   980179200
  980179200 * 18 =   463356416
  463356416 * 19 =   213837312
  213837312 * 20 =   -18221056
  -18221056 * 21 =  -382642176
 -382642176 * 22 =   171806720
  171806720 * 23 =  -343412736
 -343412736 * 24 =   348028928
  348028928 * 25 =   110788608
  110788608 * 26 = -1414463488
-1414463488 * 27 =   464191488
  464191488 * 28 =   112459776
  112459776 * 29 = -1033633792
-1033633792 * 30 =  -944242688
 -944242688 * 31 =   793247744
  793247744 * 32 =  -385875968
 -385875968 * 33 =   150994944
  150994944 * 34 =   838860800
  838860800 * 35 =  -704643072
 -704643072 * 36 =   402653184
  402653184 * 37 =  2013265920
 2013265920 * 38 =  -805306368
 -805306368 * 39 = -1342177280
-1342177280 * 40 = -2147483648
-2147483648 * 41 = -2147483648
-2147483648 * 42 =           0
          0 * 43 =           0
          0 * 44 =           0
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
          0 * 97 =           0
          0 * 98 =           0

请注意,在某些步骤中,乘法导致较小的数字(980179200 * 18 = 463356416)或不正确的符号(213837312 * 20 = -18221056),表示存在整数溢出。但是零从何而来呢?请继续阅读。

请记住,数据类型是 32 位有符号即 2 的补码整数,以下是每个步骤的说明:int

Operation         Result(1)     Binary Representation(2)                                           Result(3)
----------------  ------------  -----------------------------------------------------------------  ------------
          1 * 10            10                                                               1010            10
         10 * 11           110                                                            1101110           110
        110 * 12          1320                                                        10100101000          1320
       1320 * 13         17160                                                    100001100001000         17160
      17160 * 14        240240                                                 111010101001110000        240240
     240240 * 15       3603600                                             1101101111110010010000       3603600
    3603600 * 16      57657600                                         11011011111100100100000000      57657600
   57657600 * 17     980179200                                     111010011011000101100100000000     980179200
  980179200 * 18   17643225600                               100 00011011100111100100001000000000     463356416
  463356416 * 19    8803771904                                10 00001100101111101110011000000000     213837312
  213837312 * 20    4276746240                                   11111110111010011111100000000000     -18221056
  -18221056 * 21    -382642176  11111111111111111111111111111111 11101001001100010101100000000000    -382642176
 -382642176 * 22   -8418127872  11111111111111111111111111111110 00001010001111011001000000000000     171806720
  171806720 * 23    3951554560                                   11101011100001111111000000000000    -343412736
 -343412736 * 24   -8241905664  11111111111111111111111111111110 00010100101111101000000000000000     348028928
  348028928 * 25    8700723200                                10 00000110100110101000000000000000     110788608
  110788608 * 26    2880503808                                   10101011101100010000000000000000   -1414463488
-1414463488 * 27  -38190514176  11111111111111111111111111110111 00011011101010110000000000000000     464191488
  464191488 * 28   12997361664                                11 00000110101101000000000000000000     112459776
  112459776 * 29    3261333504                                   11000010011001000000000000000000   -1033633792
-1033633792 * 30  -31009013760  11111111111111111111111111111000 11000111101110000000000000000000    -944242688
 -944242688 * 31  -29271523328  11111111111111111111111111111001 00101111010010000000000000000000     793247744
  793247744 * 32   25383927808                               101 11101001000000000000000000000000    -385875968
 -385875968 * 33  -12733906944  11111111111111111111111111111101 00001001000000000000000000000000     150994944
  150994944 * 34    5133828096                                 1 00110010000000000000000000000000     838860800
  838860800 * 35   29360128000                               110 11010110000000000000000000000000    -704643072
 -704643072 * 36  -25367150592  11111111111111111111111111111010 00011000000000000000000000000000     402653184
  402653184 * 37   14898167808                                11 01111000000000000000000000000000    2013265920
 2013265920 * 38   76504104960                             10001 11010000000000000000000000000000    -805306368
 -805306368 * 39  -31406948352  11111111111111111111111111111000 10110000000000000000000000000000   -1342177280
-1342177280 * 40  -53687091200  11111111111111111111111111110011 10000000000000000000000000000000   -2147483648
-2147483648 * 41  -88046829568  11111111111111111111111111101011 10000000000000000000000000000000   -2147483648
-2147483648 * 42  -90194313216  11111111111111111111111111101011 00000000000000000000000000000000             0
          0 * 43             0                                                                  0             0
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
vvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvvv
          0 * 98             0                                                                  0             0
  1. 是正确的结果
  2. 是结果的内部表示(64 位用于说明)
  3. 是由两者对下32位的补码表示的结果

我们知道,将一个数字乘以一个偶数:

  • 将位向左移动,并向右添加零位
  • 得到一个偶数

因此,基本上,您的程序将一个偶数与另一个数字重复相乘,从而从右开始将结果位归零。

PS:如果乘法只涉及奇数,那么结果不会变成零。


答案 2

计算机乘法确实发生在模数2 ^ 32。一旦你在乘法中积累了足够的2次幂,那么所有值都将为0。

在这里,我们有该系列中的所有偶数,以及除以数字的2的最大幂,以及2的累积幂

num   max2  total
10    2     1
12    4     3
14    2     4
16    16    8
18    2     9
20    4    11
22    2    12
24    8    15
26    2    16
28    4    18
30    2    19
32    32   24
34    2    25
36    4    27
38    2    28
40    8    31
42    2    32

最大 42 的乘积等于 x * 2^32 = 0(mod 2^32)。两个幂的顺序与格雷码(以及其他事项)有关,并且显示为 https://oeis.org/A001511

编辑:要了解为什么对这个问题的其他回答是不完整的,请考虑这样一个事实,即同一程序,仅限于奇数整数,尽管所有溢出都不会收敛到0。