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Why should do the loop with an increasing number smaller than the first parameter in va_arg?

Time:12-16

I almost understood this function, but I wondered when I looked at the example. Why does the For statement loop until i is less than count?

int
add_em_up (int count,...)
{
  va_list ap;
  int i, sum;

  va_start (ap, count);         /* Initialize the argument list. */

  sum = 0;
  for (i = 0; i < count; i  )
    sum  = va_arg (ap, int);    /* Get the next argument value. */

  va_end (ap);                  /* Clean up. */
  return sum;
}

int
main (void)
{
  /* This call prints 16. */
  printf ("%d\n", add_em_up (3, 5, 5, 6));

  /* This call prints 55. */
  printf ("%d\n", add_em_up (10, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10));

  return 0;
}

I try to understand the concept by looking at various explanations, but it's not easy. I understand that the first parameter is the first argument of the list. It's the first parameter, so why do we have to loop around in here? But WHY? I want to know why! Help me plz!

CodePudding user response:

"Why does the For statement loop until i is less than count?"

It doesn't - it loops while i is less than count.

CodePudding user response:

Okay, this is such an easy question that experts like you must have been confused. I am implementing printf. For example, the following features are implemented:

int printf(const char *str, ...) //funtion
printf("my name : %s \n my age : %d \n", "baby coder", 1);

const char : "my name : %s \n my age : %d \n" Parameters : baby coder, 1

That's why spin the loop until less 'i' than the number of str...!

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