Function with Variable length argument

Question

I have following question on function with Variable length argument in C:

Case 1 (Works)

myPrintf("%d %s", 24, "Hi There");

Case 2 (Works)

char tempbuf[9]="Hi There";`
myPrintf("%s %d", tempbuf, 24)

Case 3 (Doesn't work)

myPrintf("%s %d", "Hi There", 24)

Does anyone has any idea why the case 3 doesn't work. I could see str = va_arg(ap, char *); returning 24 for this case intead of the actual string.

Code for myPrintf: (It is not fully functional though)

void myPrintf(char *fmt, ...)
{
int i,j,val,len;
char *str;
int len2;
va_list ap;
char tempBuf[128];

len=strlen(fmt);

memset(tempBuf,0,MAX_MSZ_LEN);

va_start(ap,fmt);

for(i=0; i<len; i++)
{
switch(fmt[i])
{
  case '%' :
  i++;
  if( fmt[i] == 's' )
  {
    str = va_arg(ap, char *);
    strcat(tempBuf, str);
  }
  else
    if( fmt[i]=='i' || fmt[i]=='d' )
    {
      val=va_arg(ap,int);
      sprintf(str,"%d",val);
      strcat(tempBuf, str);
    }
  default : 
  len2=strlen(tempBuf);
  tempBuf[len2]=fmt[i];
  }
}
va_end(ap);
}

}

We'll need to see the code for myPrintf to offer much help - it looks like a sensible code fragment so far. — Adam Wright
– Adam Wright, Commented May 24, 2011 at 15:57
What's the difference between myPrintf and myprintf for that matter? — Platinum Azure
– Platinum Azure, Commented May 24, 2011 at 15:58
Are you sure you have the comma after the format string in case 3? — Richard Pennington
– Richard Pennington, Commented May 24, 2011 at 16:04

casablanca · Accepted Answer · 2011-05-24 16:10:44Z

3

In the case for %d:

sprintf(str,"%d",val);

what does str point to? If there was a %s earlier, it points to one of the format arguments, otherwise it is uninitialized -- in both cases, it points to an invalid location for writing. You need another temporary buffer to write the value into. You were just lucky that cases 1 and 2 worked.

answered May 24, 2011 at 16:10

casablanca

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1 Comment

casablanca Over a year ago

@Punit: When %d is before %s, str is uninitialized -- you were just lucky it worked, it could have very well crashed. With %d after %s, when you used a character array for "Hi There", you could technically modify it because you allocated space for it. When you pass "Hi There" directly, it's a const string, so you can't write to it.

nos · Accepted Answer · 2011-05-24 16:19:28Z

Take a look at this code:

   if( fmt[i]=='i' || fmt[i]=='d' )
    {
      val=va_arg(ap,int);
      sprintf(str,"%d",val);
      strcat(tempBuf, str);
    }

The sprintf() call there is trying to write something to str. What is str ? When you call it like myPrintf("%s %d", "Hi There", 24) , the str will be the 2. argument, the string "Hi There". You cannot change a string literal in C, this will likely fail and might cause a crash.

When you call it like myPrintf("%s %d", tempbuf, 24), str will be tmpbuf, which is an array, which you can write to so that's fine. It only holds room for 9 bytes though, so it's easy to overflow that buffer.

You should rather just do something like

      char tmp[32];
      sprintf(tmp,"%d",val);
      strcat(tempBuf, tmp);

Richard Pennington · Accepted Answer · 2011-05-24 16:07:11Z

0

I'll go out on a limb... Put a comma after the format string for case 3.

answered May 24, 2011 at 16:07

Richard Pennington

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1 Comment

Richard Pennington Over a year ago

I'm guessing your real code is missing the comma: myPrintf("%s %d" "Hi There", 24); and the strings are getting concatenated.

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