What will be the output of the program?

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

int main()
{
    printf("%c\n", "abcdefgh"[4]);
    return 0;
}

Step 1: void fun(); This is the prototype for the function fun().

Step 2: fun(); The function fun() is called here.

The function fun() gets a character input and the input is terminated by an enter key(New line character). It prints the given character in the reverse order.

The given input characters are "abc"

Output: cba

Step 1: char *names[] = { "Suresh", "Siva", "Sona", "Baiju", "Ritu"}; The variable names is declared as an pointer to a array of strings.

Step 2: int i; The variable i is declared as an integer type.

Step 3: char *t; The variable t is declared as pointer to a string.

Step 4: t = names[3]; names[3] = names[4]; names[4] = t; These statements the swaps the 4 and 5 element of the array names.

Step 5: for(i=0; i<=4; i++) printf("%s,", names[i]); These statement prints the all the value of the array names.

Hence the output of the program is "Suresh, Siva, Sona, Ritu, Baiju".

printf("India", "CURIOUSTAB\n"); It prints "India". Because ,(comma) operator has Left to Right associativity. After printing "India", the statement got terminated.

Step 1: char str[25] = "CuriousTab"; The variable str is declared as an array of characteres and initialized with a string "CuriousTab".

Step 2: printf("%s\n", &str+2);

=> In the printf statement %s is string format specifier tells the compiler to print the string in the memory of &str+2

=> &str is a location of string "CuriousTab". Therefore &str+2 is another memory location.

Hence it prints the Garbage value.

It prints 'CuriousTab CuriousTab' in TurboC (in 16 bit platform).

It may cause a 'segmentation fault error' in GCC (32 bit platform).

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 empty".

In printf("%u %s\n", &"Hello", &"Hello");.

The %u format specifier tells the compiler to print the memory address of the "Hello1".

The %s format specifier tells the compiler to print the string "Hello2".

Hence the output of the program is "1022 Hello2".

Step 1: char *str[] = {"Frogs", "Do", "Not", "Die", "They", "Croak!"}; The variable str is declared as an pointer to the array of 6 strings.

Step 2: printf("%d, %d", sizeof(str), strlen(str[0]));

sizeof(str) denotes 6 * 4 bytes = 24 bytes. Hence it prints '24'

strlen(str[0])); becomes strlen(Frogs)). Hence it prints '5';

Hence the output of the program is 24, 5

Hint: If you run the above code in 16 bit platform (Turbo C under DOS) the output will be 12, 5. Because the pointer occupies only 2 bytes. If you run the above code in Linux (32 bit platform), the output will be 24, 5 (because the size of pointer is 4 bytes).