What will be the output of the program (in Turbo C under DOS)?

#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;
}

The file source.txt is opened in read mode and target.txt is opened in write mode. The file source.txt contains "To err is human".

Inside the while loop,

ch=getc(fs); The first character('T') of the source.txt is stored in variable ch and it's checked for EOF.

if(ch==EOF) If EOF(End of file) is true, the loop breaks and program execution stops.

If not EOF encountered, fseek(fs, 4L, SEEK_CUR); the file pointer advances 4 character from the current position. Hence the file pointer is in 5th character of file source.txt.

fputc(ch, ft); It writes the character 'T' stored in variable ch to target.txt.

The while loop runs three times and it write the character 1st and 5th and 11th characters ("Trh") in the target.txt file.

The randomize() function initializes the random number generator with a random value based on time. You can try the sample program given below in Turbo-C, it may not work as expected in other compilers.

/* Prints a random number in the range 0 to 99 */

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

int main(void)
{
    randomize();
    printf("Random number in the 0-99 range: %d\n", random (100));
    return 0;
}

The struct emp is mentioned in the prototype of the function modify() before declaring the structure.To solve this problem declare struct emp before the modify() prototype.

Step 1: int i; The variable i is declared as an integer type.

Step 1: int fun(); This prototype tells the compiler that the function fun() does not accept any arguments and it returns an integer value.

Step 1: while(i) The value of i is not initialized so this while condition is failed. So, it does not execute the while block.

Step 1: printf("Hello\n"); It prints "Hello".

Hence the output of the program is "Hello".

Given 3.14 is a double constant.

To specify 3.14 as long double, we have to add L to the 3.14. (i.e 3.14L)

A compiler may leave holes in structures by padding the first char in the structure with another byte just to ensures that the integer that follows is stored at an location. Also, there might be 2extra bytes after the integer to ensure that the long integer is stored at an address, which is multiple of 4. Such alignment is done by machines to improve the efficiency of accessing values.

In order to execute it from DOS shell, we have to run the created EXE file by entering the exe file name as C:\TC>myproc <enter>.

Step 1: int i=-3, j=2, k=0, m; here variable i, j, k, m are declared as an integer type and variable i, j, k are initialized to -3, 2, 0 respectively.

Step 2: m = ++i && ++j || ++k;
becomes m = (-2 && 3) || ++k;
becomes m = TRUE || ++k;.
(++k) is not executed because (-2 && 3) alone return TRUE.
Hence this statement becomes TRUE. So it returns '1'(one). Hence m=1.

Step 3: printf("%d, %d, %d, %d\n", i, j, k, m); In the previous step the value of i,j are increemented by '1'(one).

Hence the output is "-2, 3, 0, 1".