variadic function in C

Question

1) Why code under /* test1 */ chunk do not print out anything, but code under /* test2 */ print correctly?

2) How to use va_arg(va, char*) in /* test 1 */ code chunk.

void rdfDBG(int dbglevel, const char *fmt, ...) {

    va_list va;

#if 1 /* test 1 */
    char* logbuf;

    if ((logbuf = calloc(0x0, LOGBUFFERSIZE))== NULL)
        fprintf(stderr, "out of memory\n"); /* 1. Is there a better way instead of using fprintf? */

    va_start(va, fmt);
    (void) vsnprintf(logbuf, strlen(logbuf), fmt, va); /* 2. print nothing */
    va_end(va);

    free(logbuf);
#endif

#if 0 /* test 2 */
    va_start(va, fmt);
    vprintf(fmt, va); /* 2. print "pinfile pings6.txt" */
    va_end(va);
#endif


}


int ptInitialize(){

    char* pinfile;
    pinfile = "pings6.txt";

    rdfDBG(kINFO, "pinfile %s\n", pinfile);

    return 0;
}

Dmitri · Accepted Answer · 2011-10-14 04:55:39Z

6

The code under /* test 1 */ doesn't print anything because the vsnprint() function doesn't print to stdout; it just writes its output into the provided buffer.

It's pure luck if the code didn't crash, though, because the line:

if ((logbuf = calloc(0x0, LOGBUFFERSIZE))== NULL)

actually tells calloc() to allocate 0 bytes of memory. Because of this, I don't think it's guaranteed that the memory logbuf points at is zeroed either -- so not only is your buffer zero bytes long, but calling strlen() on it could crash or give you an invalid result.

Also, the second argument to vsnprint() is supposed to be the size of the buffer, that is the size you've allocated for logbuf, not the length of any string already in it; it's to limit the number of bytes written to the buffer to avoid a buffer overrun.

So to get everything working, you need to change:

    if ((logbuf = calloc(0x0, LOGBUFFERSIZE))== NULL)

..to..

    if ((logbuf = calloc(1, LOGBUFFERSIZE))== NULL)

..to allocate space for 1 item of LOGBUFFERSIZE bytes. And also change:

(void) vsnprintf(logbuf, strlen(logbuf), fmt, va);

..to..

vsnprintf(logbuf, LOGBUFFERSIZE, fmt, va);

..to pass it the size of your buffer and remove the useless (void) cast. And also add a line to print logbuf, such as:

fputs(logbuf, stderr);

after it.

answered Oct 14, 2011 at 4:55

Dmitri

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1 Comment

twfx Over a year ago

thanks. I have apparently misunderstood the correct usage of calloc and vsnprintf.

John Zwinck · Accepted Answer · 2011-10-14 17:01:17Z

4

This:

vsnprintf(logbuf, strlen(logbuf)...

Will never format anything, because logbuf is all zeros (having been allocated by calloc, and so strlen will return zero, which makes vsnprintf never format anything because you told it to print at most zero characters.

edited Oct 14, 2011 at 17:01

answered Oct 14, 2011 at 2:41

John Zwinck

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6 Comments

Raymond Chen Over a year ago

Even after fixing that problem, you still won't get anything printed because vsnprintf doesn't print. It puts the result into logbuf, but you never print logbuf.

John Zwinck Over a year ago

Are you out of your mind? Changing it to malloc was not the solution--now you are using uninitialized memory. You need to pass LOGBUFFERSIZE as the second argument rather than strlen(logbuf).

twfx Over a year ago

if ((logbuf = malloc(LOGBUFFERSIZE))== NULL) fprintf(stderr, "out of memory\n"); va_start(va, fmt); (void) vsnprintf(logbuf, LOGBUFFERSIZE, fmt, va); fprintf(stderr, "%s", logbuf); va_end(va);

Graeme Perrow Over a year ago

Three upvotes? Really? This answer is patently incorrect. Dmitri's answer is correct.

John Zwinck Over a year ago

How is it incorrect? I will readily admit that it is incomplete, as I see now there were actually many problems with the code, but if I said something actually wrong I would like to know what.

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