我正准备编写一个C ++函数,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(它不必自己编号itoa()
)。但后来我觉得“标准库中肯定有什么东西可以提升呢?”
注意:
答案 0 :(得分:0)
我很确定你可以在Display numbers with ordinal suffix in PHP调整四线解决方案。不幸的是,我认为在一个普通的C ++库中没有这样的东西。
答案 1 :(得分:0)
试试这个......
#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;
}
}
}
}
}
}
答案 2 :(得分:0)
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;
}
答案 3 :(得分:0)
这是我最后写的内容:
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 - 它们确实优化了简洁性,而不是其他任何东西。这更快(尽管通过将.cpp中的后缀放在函数体之外并使代码无法使用,可以更快地完成),更清晰,并且比其他大多数答案更简洁。
答案 4 :(得分:0)
// 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;
}