What will be the output of the program?

#include<stdio.h>

int main()
{
    static char mess[6][30] = {"Don't walk in front of me...", 
                               "I may not follow;", 
                               "Don't walk behind me...", 
                               "Just walk beside me...", 
                               "And be my friend." };

    printf("%c, %c\n", *(mess[2]+9), *(*(mess+2)+9));
    return 0;
}

Step 1: void swap(char *, char *); This prototype tells the compiler that the function swap accept two strings as arguments and it does not return anything.

Step 2: char *pstr[2] = {"Hello", "CuriousTab"}; The variable pstr is declared as an pointer to the array of strings. It is initialized to

pstr[0] = "Hello", pstr[1] = "CuriousTab"

Step 3: swap(pstr[0], pstr[1]); The swap function is called by "call by value". Hence it does not affect the output of the program.

If the swap function is "called by reference" it will affect the variable pstr.

Step 4: printf("%s\n%s", pstr[0], pstr[1]); It prints the value of pstr[0] and pstr[1].

Hence the output of the program is

Hello
CuriousTab

str[6] = "CURIOUSTAB"; - Nonportable pointer conversion.

The function strlen returns the number of characters in the given string.

Therefore, strlen("123456") returns 6.

Hence the output of the program is "6".

The function printf() returns the number of charecters printed on the console.

Step 1: char a[] = "\0"; The variable a is declared as an array of characters and it initialized with "\0". It denotes that the string is empty.

Step 2: if(printf("%s", a)) The printf() statement does not print anything, so it returns '0'(zero). Hence the if condition is failed.

In the else part it prints "The string is not empty".

The following examples may help you understand this problem:

1. sizeof("") returns 1 (1*).

2. sizeof("India") returns 6 (5 + 1*).

3. sizeof("CURIOUSTAB") returns 4 (3 + 1*).

4. sizeof("India\0CURIOUSTAB") returns 10 (5 + 1 + 3 + 1*).
    Here '\0' is considered as 1 char by sizeof() function.

5. sizeof("India\0CURIOUSTAB\0") returns 11 (5 + 1 + 3 + 1 + 1*).
    Here '\0' is considered as 1 char by sizeof() function.

Step 1:

printf("%d, %d, %d", sizeof(3.0f), sizeof('3'), sizeof(3.0));

The sizeof function returns the size of the given expression.

sizeof(3.0f) is a floating point constant. The size of float is 4 bytes

sizeof('3') It converts '3' in to ASCII value.. The size of int is 2 bytes

sizeof(3.0) is a double constant. The size of double is 8 bytes

Hence the output of the program is 4,2,8

Note: The above program may produce different output in other platform due to the platform dependency of C compiler.

In Turbo C, 4 2 8. But in GCC, the output will be 4 4 8.

Here str[] has declared as 7 character array and into a 8 character is stored. This will result in overwriting of the byte beyond 7 byte reserved for '\0'.

Here fputc('A', fp1); stores 'A' in the file1.c then fputc('B', fp2); overwrites the contents of the file1.c with value 'B'. Because the fp1 and fp2 opens the file1.c in write mode.

Hence the file1.c contents is 'B'.

Step 1: int k=1; The variable k is declared as an integer type and initialized to '1'.

Step 2: printf("%d == 1 is" "%s\n", k, k==1?"TRUE":"FALSE"); becomes

=> k==1?"TRUE":"FALSE"

=> 1==1?"TRUE":"FALSE"

=> "TRUE"

Therefore the output of the program is 1 == 1 is TRUE