32位无符号整数的加法

使用32个元素的数组代表unsigned int型数字(32 bit)的加法。

 1 #include <stdbool.h>
 2 #include <stdio.h>
 3 
 4 unsigned int num1[32];
 5 unsigned int num2[32];
 6 unsigned int sum[33];
 7 
 8 void init(unsigned int a, unsigned int *arr) {
 9   for (int i = 0; i < 32; ++i) {
10     arr[31 - i] = (a & 1);
11     a = a >> 1;
12   }
13 }
14 
15 void add() {
16   int s = 0;
17   int c = 0;
18   for (int i = 0; i < 32; ++i) {
19     s = num1[31 - i] + num2[31 - i] + c;
20     c = s / 2;
21     sum[32 - i] = s % 2;
22     sum[31 - i] = c;  // 数的位次(31-i)和和的进位位次(31-i)相同
23   }
24 }
25 
26 _Bool is_overflow() { return sum[0] == 1; }
27 
28 void print_num(unsigned int *arr) {
29   printf(" ");
30   for (int i = 0; i < 32; ++i) {
31     printf("%u", arr[i]);
32     if ((i + 1) % 8 == 0) {
33       printf(" ");
34     }
35   }
36   printf(".\n");
37 }
38 
39 void print_sum(unsigned int *arr) {
40   for (int i = 0; i < 33; ++i) {
41     printf("%u", arr[i]);
42     if (i > 0 && (i % 8 == 0)) {
43       printf(" ");
44     }
45   }
46   printf(".\n");
47 }
48 
49 int main() {
50   unsigned int a;
51   unsigned int b;
52   scanf("%u %u", &a, &b);
53   init(a, num1);
54   init(b, num2);
55   print_num(num1);
56   print_num(num2);
57 
58   add();
59   print_sum(sum);
60 
61   if (is_overflow()) {
62     printf("error: overflow\n");
63   } else {
64     printf("no overflow\n");
65   }
66   return 0;
67 }