我知道这是一个早期的帖子,但今天我陷入了"man"兔子洞,所以我想记录一下我的经历:
有一个名为 "strfmon()" 的函数,您可以使用 monetary.h
包含它来完成此操作,并根据本地或国际标准执行此操作。
请注意,它像 printf()
一样工作,并将采用与字符串中指定的 %
格式相同数量的 double
参数。
除了我在这里提到的内容外,还有更多内容,我发现这个页面最有帮助:https://www.gnu.org/software/libc/manual/html_node/Formatting-Numbers.html
原始答案:Original Answer
#include <monetary.h>
#include <locale.h>
#include <stdlib.h>
#include <stdio.h>
int main(){
setlocale(LC_ALL, "");
double money_amt = 1234.5678;
int buf_len = 16;
char simple_local[buf_len];
char international[buf_len];
char parenthesis_for_neg[buf_len];
char specified_width[buf_len];
char fill_6_stars[buf_len];
char fill_9_stars[buf_len];
char suppress_thousands[buf_len];
strfmon(simple_local, buf_len-1, "%n", money_amt);
strfmon(international, buf_len-1, "%i", money_amt);
strfmon(parenthesis_for_neg, buf_len-1, "%(n", money_amt);
strfmon(specified_width, buf_len-1, "%#6n", money_amt);
strfmon(fill_6_stars, buf_len-1, "%=*#6n", money_amt);
strfmon(fill_9_stars, buf_len-1, "%=*#8n", money_amt);
strfmon(suppress_thousands, buf_len-1, "%^=*#8n", money_amt);
printf( "===================== Output ===================\n"\
"Simple, local: %s\n"\
"International: %s\n"\
"parenthesis for negatives: %s\n"\
"fixed width (6 digits): %s\n"\
"fill character '*': %s\n"\
"-- note fill characters don't\n"\
"-- count where the thousdands\n"\
"-- separator would go:\n"\
"filling with 9 characters: %s\n"\
"Suppress thousands separators: %s\n"\
"================================================\n",
simple_local, international, parenthesis_for_neg,
specified_width, fill_6_stars, fill_9_stars,
suppress_thousands);
return 0;
}
===================== Output ===================
Simple, local: $1,234.57
International: USD1,234.57
parenthesis for negatives: $1,234.57
fixed width (6 digits): $ 1,234.57
fill character '*': $**1,234.57
-- note fill characters don't
-- count where the thousdands
-- separator would go:
filling with 9 characters: $*****1,234.57
Suppress thousands separators: $****1234.57
================================================