kk Exercise 6.1 Convert an integer to words

最新推荐文章于 2021-09-24 17:36:28 发布

转载最新推荐文章于 2021-09-24 17:36:28 发布 · 74 阅读

0 ·

CC 4.0 BY-SA版权

原文链接：http://www.cnblogs.com/xiaomi5320/p/4190776.html

本文提供了一个C语言程序示例，该程序接收用户输入的一个小于1000的正整数，并将其转换为英文单词形式输出。程序通过定义不同数值范围内的字符串数组来实现整数到英文单词的映射。

Exercise 6-1. write a program that will prompt for and read a positive integer less than
1,000 from the keyboard, and then create and output a string that is the value of the
integer in words. For example, if 941 is entered, the program will create the string "Nine
hundred and forty one".

//
#define __STDC_WANT_LIB_EXT1__ 1
#include <stdio.h>
#include <string.h>

int main(void)
{
  char *unit_words[] = {"zero", "one","two","three","four","five","six","seven","eight","nine"};
  char *teen_words[] = {"ten", "eleven","twelve","thirteen","fourteen","fifteen","sixteen","seventeen","eighteen","nineteen"};
  char *ten_words[] = {"error", "error","twenty","thirty","forty","fifty","sixty","seventy","eighty","ninety"};
  
  
  //上面的是什么意思？
  
  char hundred[] = " hundred";
  char and[] = " and ";         //就是这个字符。 and字符串有5+1个字符，其中一个是结束字符。
  
  char value_str[50] = "";
  int value = 0;                     // Integer to be converted
  int digits[] = {0,0,0};            // 这是什么意思?digit这个数组里面有三个值？
  int i = 0;

  printf("Enter a positive integer less than 1000: ");
  scanf_s("%d",&value);
  if(value >= 1000)
    value = 999;
  else if(value < 1)
    value = 1;//为什么不直接显示错误，而要非执行这个区间的值

  while(value > 0)
  {
    digits[i++] = value % 10;//求得各位数！
    value /= 10;//十位以上的数字
  }

  if(digits[2] > 0)
  {
    strcat_s(value_str, sizeof(value_str), unit_words[digits[2]]);// unit_WORDS[digits[2]]这是什么意思？圆程序？
    strcat_s(value_str, sizeof(value_str), hundred);
    if(digits[1] > 0 || digits[0] > 0)
      strcat_s(value_str, sizeof(value_str), and);
  }
  if(digits[1] > 0)
  {
    if(digits[1] == 1)
      strcat_s(value_str, sizeof(value_str), teen_words[digits[0]]);
    else
    {
      strcat_s(value_str,sizeof(value_str), ten_words[digits[1]]);
      if(digits[0] > 0)
      {
        strcat_s(value_str, sizeof(value_str), " ");
        strcat_s(value_str, sizeof(value_str), unit_words[digits[0]]);
      }
    }
  }
  else
    if(digits[0] > 0)
      strcat_s(value_str, sizeof(value_str), unit_words[digits[0]]);
      
    //最终一步到位的输出，程序相当简洁，上面的构思巧妙
  printf("\n%s\n", value_str);
  return 0;
}

转载于:https://www.cnblogs.com/xiaomi5320/p/4190776.html