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。

 

#include <stdbool.h>
#include <stdio.h>

unsigned int num1[32];
unsigned int num2[32];
unsigned int sum[33];

void init(unsigned int a, unsigned int *arr) {
  for (int i = 0; i < 32; ++i) {
    arr[31 - i] = (a & 1);
    a = a >> 1;
  }
}

void add() {
  int s = 0;
  int c = 0;
  for (int i = 0; i < 32; ++i) {
    s = num1[31 - i] + num2[31 - i] + c;
    c = s / 2;
    sum[32 - i] = s % 2;
    sum[31 - i] = c;  // 数的位次(31-i)和和的进位位次(31-i)相同
  }
}

_Bool is_overflow() { return sum[0] == 1; }

void print_num(unsigned int *arr) {
  printf(" ");
  for (int i = 0; i < 32; ++i) {
    printf("%u", arr[i]);
    if ((i + 1) % 8 == 0) {
      printf(" ");
    }
  }
  printf(".\n");
}

void print_sum(unsigned int *arr) {
  for (int i = 0; i < 33; ++i) {
    printf("%u", arr[i]);
    if (i > 0 && (i % 8 == 0)) {
      printf(" ");
    }
  }
  printf(".\n");
}

int main() {
  unsigned int a;
  unsigned int b;
  scanf("%u %u", &a, &b);
  init(a, num1);
  init(b, num2);
  print_num(num1);
  print_num(num2);

  add();
  print_sum(sum);

  if (is_overflow()) {
    printf("error: overflow\n");
  } else {
    printf("no overflow\n");
  }
  return 0;
}