th
在显示“ ”中的“”等文本时,我需要一个函数来返回几天的后缀Wednesday June 5th, 2008
。
它只需要处理数字 1 到 31(不需要错误检查)和英语。
这是一个替代方案,它也适用于更大的数字:
static const char *daySuffixLookup[] = { "th","st","nd","rd","th",
"th","th","th","th","th" };
const char *daySuffix(int n)
{
if(n % 100 >= 11 && n % 100 <= 13)
return "th";
return daySuffixLookup[n % 10];
}
以下函数适用于 C:
char *makeDaySuffix (unsigned int day) {
//if ((day < 1) || (day > 31)) return "";
switch (day) {
case 1: case 21: case 31: return "st";
case 2: case 22: return "nd";
case 3: case 23: return "rd";
}
return "th";
}
根据要求,它仅适用于数字 1 到 31(含)。如果您想要(可能但不一定)原始速度,您可以尝试:
char *makeDaySuffix (unsigned int day) {
static const char * const suffix[] = {
"st","nd","rd","th","th","th","th","th","th","th",
"th","th","th","th","th","th","th","th","th","th"
"st","nd","rd","th","th","th","th","th","th","th"
"st"
};
//if ((day < 1) || (day > 31)) return "";
return suffix[day-1];
}
您会注意到,尽管已注释掉,但我有边界检查。如果传入一个意外值的可能性很小,您可能需要取消注释这些行。
请记住,对于当今的编译器,关于在高级语言中什么是更快的天真假设可能是不正确的:测量,不要猜测。
const char *getDaySuffix(int day) {
if (day%100 > 10 && day%100 < 14)
return "th";
switch (day%10) {
case 1: return "st";
case 2: return "nd";
case 3: return "rd";
default: return "th";
};
}
这个适用于任何数字,而不仅仅是 1-31。
在这里查看我的问题:如何将基数转换为序数(它不是 C# 的)。
摘要:看起来还没有办法,由于您的要求有限,您可以使用像发布的那样的简单功能。