如何在C中将数字从1123456789格式化为1,123,456,789？

Question

如何用C语言将数字从格式化1123456789为1,123,456,789？我尝试使用，printf("%'10d\n", 1123456789);但这不起作用。

你能建议什么吗？解决方案越简单越好。

Answer 1

如果您的printf支持该'标志（POSIX 2008要求printf()），则可以仅通过适当地设置区域设置来实现。例：#include <stdio.h>#include <locale.h>int main(void){&nbsp; &nbsp; setlocale(LC_NUMERIC, "");&nbsp; &nbsp; printf("%'d\n", 1123456789);&nbsp; &nbsp; return 0;}并运行：$ ./example&nbsp;1,123,456,789在Mac OS X和Linux（Ubuntu 10.10）上进行了测试。

Answer 2

您可以按以下方式递归执行此操作（请注意，INT_MIN如果使用二进制补码，则需要额外的代码来管理它）：void printfcomma2 (int n) {&nbsp; &nbsp; if (n < 1000) {&nbsp; &nbsp; &nbsp; &nbsp; printf ("%d", n);&nbsp; &nbsp; &nbsp; &nbsp; return;&nbsp; &nbsp; }&nbsp; &nbsp; printfcomma2 (n/1000);&nbsp; &nbsp; printf (",%03d", n%1000);}void printfcomma (int n) {&nbsp; &nbsp; if (n < 0) {&nbsp; &nbsp; &nbsp; &nbsp; printf ("-");&nbsp; &nbsp; &nbsp; &nbsp; n = -n;&nbsp; &nbsp; }&nbsp; &nbsp; printfcomma2 (n);}总结：用户printfcomma使用整数调用，负数的特殊情况是通过简单地打印“-”并使数字为正数来处理（这是不能使用的位INT_MIN）。输入时printfcomma2，小于1,000的数字将被打印并返回。否则，递归将在下一级上调用（因此将调用1,234,567，先以1,234，然后是1），直到找到小于1,000的数字。然后将打印该数字，我们将返回递归树，在打印过程中打印逗号和下一个数字。还有一个更简洁的版本，尽管它在检查每个级别的负数时进行了不必要的处理（这并不重要，因为递归级别的数量有限）。这是一个完整的测试程序：#include <stdio.h>void printfcomma (int n) {&nbsp; &nbsp; if (n < 0) {&nbsp; &nbsp; &nbsp; &nbsp; printf ("-");&nbsp; &nbsp; &nbsp; &nbsp; printfcomma (-n);&nbsp; &nbsp; &nbsp; &nbsp; return;&nbsp; &nbsp; }&nbsp; &nbsp; if (n < 1000) {&nbsp; &nbsp; &nbsp; &nbsp; printf ("%d", n);&nbsp; &nbsp; &nbsp; &nbsp; return;&nbsp; &nbsp; }&nbsp; &nbsp; printfcomma (n/1000);&nbsp; &nbsp; printf (",%03d", n%1000);}int main (void) {&nbsp; &nbsp; int x[] = {-1234567890, -123456, -12345, -1000, -999, -1,&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;0, 1, 999, 1000, 12345, 123456, 1234567890};&nbsp; &nbsp; int *px = x;&nbsp; &nbsp; while (px != &(x[sizeof(x)/sizeof(*x)])) {&nbsp; &nbsp; &nbsp; &nbsp; printf ("%-15d: ", *px);&nbsp; &nbsp; &nbsp; &nbsp; printfcomma (*px);&nbsp; &nbsp; &nbsp; &nbsp; printf ("\n");&nbsp; &nbsp; &nbsp; &nbsp; px++;&nbsp; &nbsp; }&nbsp; &nbsp; return 0;}输出为：-1234567890&nbsp; &nbsp; : -1,234,567,890-123456&nbsp; &nbsp; &nbsp; &nbsp; : -123,456-12345&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;: -12,345-1000&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; : -1,000-999&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;: -999-1&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;: -10&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; : 01&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; : 1999&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; : 9991000&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp;: 1,00012345&nbsp; &nbsp; &nbsp; &nbsp; &nbsp; : 12,345123456&nbsp; &nbsp; &nbsp; &nbsp; &nbsp;: 123,4561234567890&nbsp; &nbsp; &nbsp;: 1,234,567,890一个不信任递归的迭代解决方案（尽管递归的唯一问题往往是堆栈空间，这在这里不会成为问题，因为即使对于64位整数也只有几层深度）：void printfcomma (int n) {&nbsp; &nbsp; int n2 = 0;&nbsp; &nbsp; int scale = 1;&nbsp; &nbsp; if (n < 0) {&nbsp; &nbsp; &nbsp; &nbsp; printf ("-");&nbsp; &nbsp; &nbsp; &nbsp; n = -n;&nbsp; &nbsp; }&nbsp; &nbsp; while (n >= 1000) {&nbsp; &nbsp; &nbsp; &nbsp; n2 = n2 + scale * (n % 1000);&nbsp; &nbsp; &nbsp; &nbsp; n /= 1000;&nbsp; &nbsp; &nbsp; &nbsp; scale *= 1000;&nbsp; &nbsp; }&nbsp; &nbsp; printf ("%d", n);&nbsp; &nbsp; while (scale != 1) {&nbsp; &nbsp; &nbsp; &nbsp; scale /= 1000;&nbsp; &nbsp; &nbsp; &nbsp; n = n2 / scale;&nbsp; &nbsp; &nbsp; &nbsp; n2 = n2&nbsp; % scale;&nbsp; &nbsp; &nbsp; &nbsp; printf (",%03d", n);&nbsp; &nbsp; }}这两个产生2,147,483,647的INT_MAX。