Thread: how to add a parameter to the function

no you cannot do it, you should also pass the additional parameter...

What you can do - to write a function with another name and additional parameter
and replace the body of the old function with the call to the new one

Then all calls to the old one can be leaved untoched, and new places will call the new function

Code:

int doWork(int a, int b) //old function
{
  return doWorkNew(a,b,10 /*default value for the new paraameter*/);
}

int doWorkNew(int a, int b, int c)
{
//copy here the body of the old doWork function and modify it to process the new parameter
}

Originally Posted by andor

Asume that you define function like this:

Code:

int foo(void *);

than you can pass to function different number of arguments. Okay its only one real argument but void * can point to array of arguments.

Code:

#include <stdio.h>

int foo(void *);

int main()
{
   char arg1[] = {2, 1, 2};
   char arg2[] = {3, 1, 2, 3};

   printf("%d\n", foo((void *) arg1));
   printf("%d\n", foo((void *) arg2));
  
  return 0;
}

int foo(void * arg)
{
   int acc=0, i;
   char * tmp = arg;
   
/* first element of array is num of arguments */   
   for (i=0; i<tmp[0]; i++)
   {
      acc += tmp[i+1];
   }
   
   return acc;
}

int this case better to pass the char* pointer without cast

No you cannot pass a different number of arguments. You're stil passing one argument. You're passing an array. That's one argument. Void pointers don't help you get around that.


Quzah.