C: Ellipsis operator (…) : printf

Ever imagined how printf works, even though we are able to pass a number of arguments to it. If we design a function which takes two arguments and pass three parameters, we are bound to get this error “too many arguments to function”i.e., suppose we have a function
    int fun2(int a, int b)

and we call the function

    fun2(2,3,4)

we are sure to get the above error. So the question is how printf / scanf works with variable number of arguments? This is because C has a feature called ellipsis (…) by which you are able to pass variable number of arguments?

So the prototype of printf is

    int printf(const char *str,...)

But the next question is how then 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

Below is a simple code which shows how this can be done

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);

        print("3",2,3,4);

}
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