Passing and using variable number of arguments to function in C

Question

I don't understand why this doesn't print out "this is a test 42" like I'm expecting it to?

  1 #include <stdio.h>
  2 #include <stdarg.h>
  3 
  4 #define ME(x)   blah x
  5 
  6 void blah(const char *fmt, ...)
  7 {
  8         va_list arg;
  9 
 10         va_start(arg, fmt);
 11         printf(fmt, arg);
 12         va_end(arg);
 13 }
 14 
 15 int main()
 16 {
 17         ME(("this is a test %d\n", 42));
 18 
 19         return 0;
 20 }

Instead it's something like this:

$ gcc blah.c
$ ./a.out
this is a test 1606416656

Uli Schlachter · Accepted Answer · 2010-09-01 07:24:13Z

8

You want to call vprintf() instead of printf().

answered Sep 1, 2010 at 7:24

Uli Schlachter

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2 Comments

Brian Over a year ago

Thank you very much! In the linux kernel, I'm assuming there is also vprintk?

Uli Schlachter Over a year ago

Seems so, googling for "kernel vprintk" found this: lkml.indiana.edu/hypermail/linux/kernel/0407.3/1688.html

Patrick · Accepted Answer · 2010-09-01 07:25:53Z

3

You should use va_arg to get the actual argument value. Va_start is only an initialization of the arg variable. Arg is actualy a pointer to the value on the stack, it's not the valut itself.

The following line gets the actual value:

int myvalue = va_arg(arg,int);

Notice that I get an int and not a short, since short's are automatically promoted to int by the C compiler.

EDIT: Uli's answer is also correct. If you want to pass multiple values to printf, you should call vprintf instead of printf (and then calling va_arg is not needed, since in this case you don't know the exact types of the arguments).

answered Sep 1, 2010 at 7:25

Patrick

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