获取数字序列的功能

Function for getting the ordinal of a number?

本文关键字：功能 字序 数字 获取      更新时间：2023-10-16

i即将编写一个C 函数来执行以下操作：

1 ---> "1st"
2 ---> "1nd"
3 ---> "3rd"
...
17657 --> "17657th"
...

即。生成该数字的序数扩展字符串（它不必执行数字本身的itoa()）。但是后来我以为"标准库中的某些东西或提升已经做到了吗？"

注意：

  • 我知道写这篇文章并不难，这里有一个实现，因此我只是不想复制代码。
  • 显然，我需要英语。多语言版本对政治校正的考虑是很好的，而不仅仅是……

这是我最终写的：

const char* ordinal_suffix(int n)
{
        static const char suffixes [][3] = {"th", "st", "nd", "rd"};
        auto ord = n % 100;
        if (ord / 10 == 1) { ord = 0; }
        ord = ord % 10;
        if (ord > 3) { ord = 0; }
        return suffixes[ord];
}

代码高尔夫解决方案很可爱，但是 - 它们确实为简洁而不是其他任何东西都进行了优化。这是更快的（尽管可以通过将后缀放在功能主体中并使代码不限于代码来使其更快，但比这里大多数其他答案更清晰，更简短。

// Returns numbers with ordinal suffix as string
// Based on https://stackoverflow.com/questions/3109978/display-numbers-with-ordinal-suffix-in-php
std::string NumberToOrdinal(size_t number) {
  std::string suffix = "th";
  if (number % 100 < 11 || number % 100 > 13) {
    switch (number % 10) {
      case 1:
        suffix = "st";
        break;
      case 2:
        suffix = "nd";
        break;
      case 3:
        suffix = "rd";
        break;
    }
  }
  return std::to_string(number) + suffix;
}

我敢肯定，您可以在PHP中使用Ordinal后缀以显示四行解决方案。不幸的是，我认为在常见的C lib中没有这样的事情。

尝试这个... 

#include <iostream>
using namespace std;
void suffix(int n, char suff[]);
// creates the ordinal suffix
// for a given number
int main()
{
  char s[5];
  int x;
  cout << "Enter a number to find the ordinal suffix for ";
  cin >> x;
  suffix(52111,s);
}
void suffix(int n, char suff[])
{
   if(n%100 == 11 || n%100 == 12 || n%100 == 13)
    {
      cout << "suffix is: " << n << "th";
      cout << endl;
    }
  else
    {
      if(n%10 == 1)
    {
      cout << "Suffix is: " << n << "st";
      cout << endl;
    }
      else
    {
      if(n%10 == 2)
        {
          cout << "Suffix is: " << n << "nd";
          cout << endl;
        }
      else
        {
          if(n%10 == 3)
        {
          cout << "Suffix is: " << n << "rd";
          cout << endl;
        }
          else
        {
          if(n%10 == 4 || n%10 == 5 || n%10 == 6 || n%10 == 7 || n%10 == 8 || n%10 == 9 || n%10 == 0)
              {
            cout << "Suffix is: " << n << "th";
            cout << endl;
              }
        }
        }
    }
    }
}

我使用以下字符串函数来完成它。

#include <string>
#include <iostream>
using namespace std;
string ordinal(int i)
{
    if(i==1)
    { 
        return "First";
    }
    if(i==2)
    { 
        return "Second";
    }
    if(i==3)
    { 
        return "Third";
    }
    if(i==4)
    { 
        return "Fourth";
    }
    if(i==5)
    { 
        return "Fifth";
    }
    if(i==6)
    { 
        return "Sixth";
    }
    if(i==7)
    { 
        return "Seventh";
    }
    if(i==8)
    { 
        return "Eighth";
    }
}
int main()
{
    for(int i=0; i<8; i++) 
    {
        cout << ordinal(i+1) << " number: ";
    }
    return 0;
}


#include <iostream>
#include <string>
std::string number_to_ordinal(int number)
{
  // Convert number to string
  std::string ordinal = std::to_string(number);
  
  // Get the last character of the number to later determine ordinal indicator
  char last_char = ordinal.back();
  
  // Get the last two characters of the number to deal with ordinal indicator conditions
  std::string last_two_char;
  if(ordinal.size() > 1)
    last_two_char = ordinal.substr(ordinal.size() - 2, ordinal.size());
    
  // Determine ordinal indicator. Each number with a last character ending in '1', '2',
  // and '3' require ordinal indicators of 'st', 'nd', and 'rd', respectively. However,
  // numbers with the last two characters ending in '11', '12', and '13' require 'th' 
  // as the ordinal indicator.
  if(last_two_char != "11" && last_char == '1')
    ordinal += "st";
  else if (last_two_char != "12" && last_char == '2')
    ordinal += "nd";
  else if (last_two_char != "13" && last_char == '3')
    ordinal +="rd";
  else
    ordinal += "th"; // All other numbers require 'th' as the ordinal indicator
    
  return ordinal;
}
///////////////////////////////////////////////////////////////////////
//  Main Program
///////////////////////////////////////////////////////////////////////
int main()
{
  // Test number to ordinal
  for(int i = 17657; i < 17725; i++)
    std::cout << number_to_ordinal(i) << std::endl;
  return 0;
}

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