C Language Variable arguments Using a format string
Example
Using a format string provides information about the expected number and type of the subsequent variadic arguments in such a way as to avoid the need for an explicit count argument or a terminator value.
The example below shows a a function that wraps the standard printf() function, only allowing for the use of variadic arguments of the type char, int and double (in decimal floating point format). Here, like with printf(), the first argument to the wrapping function is the format string. As the format string is parsed the function is able to determine if there is another variadic argument expected and what it's type should be.
#include <stdio.h>
#include <stdarg.h>
int simple_printf(const char *format, ...)
{
va_list ap; /* hold information about the variadic argument list. */
int printed = 0; /* count of printed characters */
va_start(ap, format); /* start variadic argument processing */
while (*format != '\0') /* read format string until string terminator */
{
int f = 0;
if (*format == '%')
{
++format;
switch(*format)
{
case 'c' :
f = printf("%d", va_arg(ap, int)); /* print next variadic argument, note type promotion from char to int */
break;
case 'd' :
f = printf("%d", va_arg(ap, int)); /* print next variadic argument */
break;
case 'f' :
f = printf("%f", va_arg(ap, double)); /* print next variadic argument */
break;
default :
f = -1; /* invalid format specifier */
break;
}
}
else
{
f = printf("%c", *format); /* print any other characters */
}
if (f < 0) /* check for errors */
{
printed = f;
break;
}
else
{
printed += f;
}
++format; /* move on to next character in string */
}
va_end(ap); /* end variadic argument processing */
return printed;
}
int main (int argc, char *argv[])
{
int x = 40;
int y = 0;
y = simple_printf("There are %d characters in this sentence", x);
simple_printf("\n%d were printed\n", y);
}
