C: Passing variable number of arguments from one function to another

Question

So, here's a small problem I'm facing right now -> I'm trying to write a function that will accept a char* message and a variable number of arguments. My function will modify the message a little, and then It'll call printf with the message and given parameters. Essentialy, I'm trying to write something like that:

void modifyAndPrintMessage(char* message,...){
    char* newMessage; //copy message.
    //Here I'm modifying the newMessage to be printed,and then I'd like to print it. 
    //passed args won't be changed in any way.

    printf(newMessage,...); //Of course, this won't work. Any ideas?
    fflush(stdout);

}

So, anybody knows what should I do to make it happen? I'd be most grateful for any help :)

http://bobobobo.wordpress.com/2008/01/28/how-to-use-variable-argument-lists-va_list/ — frickskit, Apr 05 '13 at 14:30
This is not really a duplicate of [SO 15830641](http://stackoverflow.com/q/15830641), nor of [SO 15836392](http://stackoverflow.com/q/15836392) which the other is closed as a duplicate of. — Jonathan Leffler, Apr 05 '13 at 22:35

K Scott Piel · Answer 1 · 2013-04-05T14:52:33.010

17

You want to use varargs...

void modifyAndPrintMessage( char* message, ... )
{
    // do somehthing custom

    va_list args;
    va_start( args, message );

    vprintf( newMessage, args );

    va_end( args );
}

edited Apr 05 '13 at 14:52

answered Apr 05 '13 at 14:30

K Scott Piel

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I wasn't aware that such a thing as vprintf existed! Thanks! – JJS Apr 05 '13 at 14:39
@IanTraum, shouldn't the code be something like `va_start(args, message);` followed by `vprintf(message, args);`? – Edward Clements Apr 05 '13 at 14:47
@Edward Clements -- you are correct... copy/paste error on my part. I fixed it. – K Scott Piel Apr 05 '13 at 14:53

Edward Clements · Answer 2 · 2013-09-01T17:26:56.367

4

void modifyAndPrintMessage(char* message,...)
{   char newMessage[1024]; // **Make sure the buffer is large enough**
    va_list args;
    va_start(args, message);
    vsnprintf(newMessage, message, args);
    printf(newMessage);
    fflush(stdout);
}

edited Sep 01 '13 at 17:26

answered Apr 05 '13 at 14:30

Edward Clements

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K Scott Piel gave an even simpler way of doing this, by using vprintf instead of vsnprintf, but your way is also good and clean. Thank you! – JJS Apr 05 '13 at 14:41
yes, but in that case shouldn't the code be something like `va_start(args, message);` followed by `vprintf(message, args);`? – Edward Clements Apr 05 '13 at 14:46
1. `printf("%s", newMessage)` - Otherwise you expand `%` twice. 2. `newMessage[sizeof(newMessage)-1] = '\0'` will assure termination (so long messages will be just truncated). – ugoren Aug 29 '13 at 08:37
And you need `va_end`. On most platforms it does nothing, but still needed. – ugoren Aug 29 '13 at 08:40

score 0 · Answer 3 · answered Apr 05 '13 at 14:27

0

Use varargs to accept variable number of parameters then use sprintf to create the new message

answered Apr 05 '13 at 14:27

msam

4,259
3
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32

1

Actually **v**sprintf should be used. – Artem Pisarenko Nov 18 '15 at 05:01

Paul · Answer 4 · 2013-04-06T08:53:44.260

You can use va_list from stdarg.h,

C example: http://www.tutorialspoint.com/cprogramming/c_variable_arguments.htm C++ example: http://www.cprogramming.com/tutorial/lesson17.html.

An of course, see the man page: http://linux.die.net/man/3/stdarg

Man page example for reference:

#include <stdio.h>

#include <stdarg.h>

void
foo(char *fmt, ...)
{
    va_list ap;
    int d;
    char c, *s;

   va_start(ap, fmt);
    while (*fmt)
        switch (*fmt++) {
        case 's':              /* string */
            s = va_arg(ap, char *);
            printf("string %s\n", s);
            break;
        case 'd':              /* int */
            d = va_arg(ap, int);
            printf("int %d\n", d);
            break;
        case 'c':              /* char */
            /* need a cast here since va_arg only
               takes fully promoted types */
            c = (char) va_arg(ap, int);
            printf("char %c\n", c);
            break;
        }
    va_end(ap);
}

This is a C question; `` is a C++ header; `` is a non-standard header (or typo). — Jonathan Leffler, Apr 05 '13 at 22:36

score -2 · Answer 5 · answered Apr 05 '13 at 14:30

There is a library which includes this functionality. Here is some example code from the reference:

#include <stdarg.h>     /* va_list, va_start, va_arg, va_end */

int FindMax (int n, ...)
{
    int i,val,largest;
    va_list vl;
    va_start(vl,n);
    largest=va_arg(vl,int);
    for (i=1;i<n;i++)
    {
        val=va_arg(vl,int);
        largest=(largest>val)?largest:val;
    }
    va_end(vl);
    return largest;
}

The ellipsis is actually valid code, and you can use the va_list object to parse a variable number of parameters.

Thank you for answering, but I know what varargs are, and that wasn't what I asked for. — JJS, Apr 05 '13 at 14:42

C: Passing variable number of arguments from one function to another

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