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);
}

}

Answer 1

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.