What will be the output of the program (myprog.c) given below if it is executed from the command line?
cmd> myprog one two three

/* myprog.c */
#include<stdio.h>
#include<stdlib.h>

int main(int argc, char **argv)
{
    printf("%s\n", *++argv);
    return 0;
}

Step 1: int i=4, j=-1, k=0, w, x, y, z; here variable i, j, k, w, x, y, z are declared as an integer type and the variable i, j, k are initialized to 4, -1, 0 respectively.

Step 2: w = i || j || k; becomes w = 4 || -1 || 0;. Hence it returns TRUE. So, w=1

Step 3: x = i && j && k; becomes x = 4 && -1 && 0; Hence it returns FALSE. So, x=0

Step 4: y = i || j &&k; becomes y = 4 || -1 && 0; Hence it returns TRUE. So, y=1

Step 5: z = i && j || k; becomes z = 4 && -1 || 0; Hence it returns TRUE. So, z=1.

Step 6: printf("%d, %d, %d, %d\n", w, x, y, z); Hence the output is "1, 0, 1, 1".

strrchr() returns a pointer to the last occurrence of character in a string.

Example:

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

int main()
{
    char str[30] = "12345678910111213";
    printf("The last position of '2' is %d.\n",
            strrchr(str, '2') - str);
    return 0;
}

Output: The last position of '2' is 14.

FALSE, The macro is case insensitive.

Here the type name mystruct is known at the point of declaring the structure, as it is already defined.

near=2, far=4 and huge=4 pointers exist only under DOS. Under windows and Linux every pointers is 4 bytes long.

A structure pointing to itself is called self-referential structures.

Step 1: int arr[]={2, 3, 4, 1, 6}; The variable arr is declared as an integer array and initialized.

Step 2: printf("%u, %u, %u\n", arr, &arr[0], &arr); Here,

The base address of the array is 1200.

=> arr, &arr is pointing to the base address of the array arr.

=> &arr[0] is pointing to the address of the first element array arr. (ie. base address)

Hence the output of the program is 1200, 1200, 1200

The system will allocate 2 bytes for the union.

The statements u.ch[0]=3; u.ch[1]=2; store data in memory as given below.