Does there exist any way to make the command-line arguments available to other functions without passing them as arguments to the function?

What will be the output of the program?

#include<stdio.h>
int get();

int main()
{
    const int x = get();
    printf("%d", x);
    return 0;
}
int get()
{
    return 20;
}

Point out the error in the program.

#include<stdio.h>
#include<stdlib.h>
int fun(const union employee *e);

union employee
{
    char name[15];
    int age;
    float salary;
};
const union employee e1;

int main()
{
    strcpy(e1.name, "A");
    fun(&e1);
    printf("%s %d %f", e1.name, e1.age, e1.salary);
    return 0;
}
int fun(const union employee *e)
{
    strcpy((*e).name, "B");
    return 0;
}

What function should be used to free the memory allocated by calloc() ?

Point out the correct statement which correctly free the memory pointed to by 's' and 'p' in the following program?

#include<stdio.h>
#include<stdlib.h>

int main()
{
    struct ex
    {
        int i;
        float j;
        char *s
    };
    struct ex *p;
    p = (struct ex *)malloc(sizeof(struct ex));
    p->s = (char*)malloc(20);
    return 0;
}

What do the following declaration signify?

void *cmp();

What will be the output of the program?

#include<stdio.h>

int main()
{
    char huge *near *far *ptr1;
    char near *far *huge *ptr2;
    char far *huge *near *ptr3;
    printf("%d, %d, %d\n", sizeof(**ptr1), sizeof(ptr2), sizeof(*ptr3));
    return 0;
}

Point out the error in the following program.

#include<stdio.h>
void display(int (*ff)());

int main()
{
    int show();
    int (*f)();
    f = show;
    display(f);
    return 0;
}
void display(int (*ff)())
{
    (*ff)();
}
int show()
{
    printf("IndiaBIX");
}

We can modify the pointers "source" as well as "target".

Will the program outputs "IndiaBIX.com"?

#include<stdio.h>
#include<string.h>

int main()
{
    char str1[] = "IndiaBIX.com";
    char str2[20];
    strncpy(str2, str1, 8);
    printf("%s", str2);
    return 0;
}