What will be the output of the program?

#include<stdio.h>
int main()
{
    int i=3;
    switch(i)
    {
        case 1:
            printf("Hello\n");
        case 2:
            printf("Hi\n");
        case 3:
            continue;
        default:
            printf("Bye\n");
    }
    return 0;
}

Step 1: char ch = 'A'; The variable ch is declared as an character type and initialized with value 'A'.

Step 2:

printf("%d, %d, %d", sizeof(ch), sizeof('A'), sizeof(3.14));

The sizeof function returns the size of the given expression.

sizeof(ch) becomes sizeof(char). The size of char is 1 byte.

sizeof('A') becomes sizeof(65). The size of int is 4 bytes (as mentioned in the question).

sizeof(3.14f). The size of float is 4 bytes.

Hence the output of the program is 1, 4, 4

Step 1: char *p1 = "India", *p2; The variable p1 and p2 is declared as an pointer to a character value and p1 is assigned with a value "India".

Step 2: p2=p1; The value of p1 is assigned to variable p2. So p2 contains "India".

Step 3: p1 = "CURIOUSTAB"; The p1 is assigned with a string "CURIOUSTAB"

Step 4: printf("%s %s\n", p1, p2); It prints the value of p1 and p2.

Hence the output of the program is "CURIOUSTAB India".

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.

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".

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

The program will generate infinite loop. When an EOF is encountered fgetc() returns EOF. Instead of checking the condition for EOF we have checked it for NULL. so the program will generate infinite loop.

float a=3.15529; The variable a is declared as an float data type and initialized to value 3.15529;

printf("%2.1f\n", a); The precision specifier tells .1f tells the printf function to place only one number after the .(dot).

Hence the output is 3.2

int a=250; The variable a is declared as an integer type and initialized to value 250.

printf("%1d\n", a); It prints the value of variable a.

Hence the output of the program is 250.