What is Ellipsis ?
Ans:
Suppose we declare some prototypes:
a) int factorial(int);
b) int sum(int,int);
Then the function calls should be somewhat like this :
A) factorial(n);
B) sum(a,b);
But if we write in the following way :
factorial(n1,n2);
sum(a,c,b);
Then we will definitely encounter an error.
That's because we have already declared the number of
arguments which the respective functions can take. So ,
we cannot put more arguments than the specified ones.
But if this happens in the functions in C, then what about
the functions "printf()" and "scanf()". We can take any
number of arguments according to our choice. So, what
do their prototypes look like ???
THey look somewhat like this :
int printf(const int *str, ...)
Now here's where the "ELLIPSIS" comes into play. This
ellipsis assures the entry of any number of argument
inputs or output as desired by the user.
***********************************************
After this, another question arises , how can we access
the arguments in the function. ??
This can be done by the power of pointers. Let’s take a
pointer which points to the last argument before …
and depending on the next arguments of what we expect,
we increment the pointer and increment it accordingly.
eg:
#include<stdio.h>
#include<conio.h>
#include<stdlib.h>
int print(const char *str,...)
/*str has the number of integers passed*/
{
int i;
int num_count=atoi(str);
int *num=(int *)&str;
for(i=1;i<=num_count;i++)
printf("%d ",*(num+i));}
int print_num(int num_count,...)
/*num_count contains the number of integers passed*/
{
int i;
int *num=&num_count;
for(i=1;i<=num_count;i++)
printf("%d ",*(num+i));
}
int main()
{
print_num(3,2,3,4);
printf("\n");
print("3",2,3,4);
}
