2008 November 24th Monday (十一月　二十四日　月曜日)

I almost forgot how to implement a vararg function in C programming language. There is a example.

#include <stdarg.h>

/* minprintf: minimal printf with variable argument list */
void minprintf(char *fmt, ...)
{
va_list ap; /* points to each unnamed arg in turn */
char *p, *sval;
int ival;
double dval;

va_start(ap, fmt); /* make ap point to 1st unnamed arg */
for (p = fmt; *p; p++) {
if (*p != '%') {
putchar(*p);
continue;
}
switch (*++p) {
case 'd':
ival = va_arg(ap, int);
printf("%d", ival);
break;
case 'f':
dval = va_arg(ap, double);
printf("%f", dval);
break;
case 's':
for (sval = va_arg(ap, char *); *sval; sval++)
putchar(*sval);
break;
default:
putchar(*p);
break;
}
}
va_end(ap); /* clean up when done */
}

Another example is here.

#include<stdio.h>
#include<stdarg.h>
int my_printf( const char *format,……)
{
va_list ap;
int retval;
va_start(ap,format);

printf(“my_printf( ):”);
retval = vprintf(format,ap);
va_end(ap);

return retval;
}

main()
{
int i = 150,j = -100;
double k = 3.14159;

my_printf(“%d %f %x/n”,j,k,i);
my_printf(“%2d %*d/n”,i,2,i);
}

run my_printf() : -100 3.14159 96
my_printf() : 150 150