Question

I have a function

void func(int x, char *str, ...)
{
  ...
}

I am invoking it as follows:

func(1, "1", "2", "3");

How can I print the values of all the extra arguments (2, 3) in function?

Answer 1

From the man page of STDARG about the use of va_arg to get the next argument:

If there is no next argument, or
if type is not compatible with the
type of the actual next argument (as
promoted according to the default
argument promotions), random errors
will occur.

Hence, unless you want random errors to creep in, you should know the number of arguments beforehand.

Even so, if you want to throw caution to the winds, you could try:

void func(int x,char *str, ...)
{
    va_list al;
    va_start(al,str);

    while(x>0)
    {
        str=va_arg(al,char *);
        --x;      
    }

    while(str != NULL)
    {
        printf("%s ",str);
        str=va_arg(al,char *);
    }
    va_end(al);
}

With,

func(1,"1","2","3");

I got the output,

2 3 U��WVS�O  

If it satisfies your purpose, you could pick out the required number of arguments from this output.