What will be the output of the program?

#include<stdio.h>

int main()
{
    printf(5+"Good Morning\n");
    return 0;
}

Step 1: char p[] = "%d\n"; The variable p is declared as an array of characters and initialized with string "%d".

Step 2: p[1] = 'c'; Here, we overwrite the second element of array p by 'c'. So array p becomes "%c".

Step 3: printf(p, 65); becomes printf("%c", 65);

Therefore it prints the ASCII value of 65. The output is 'A'.

The statement str = "Kanpur"; generates the LVALUE required error. We have to use strcpy function to copy a string.

To remove error we have to change this statement str = "Kanpur"; to strcpy(str, "Kanpur");

The program prints the string "anpur"

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).

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".

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".

The line char str = "CuriousTab"; generates "Non portable pointer conversion" error.

To eliminate the error, we have to change the above line to

char *str = "CuriousTab"; (or) char str[] = "CuriousTab";

Then it prints "CuriousTab".

printf(5+"CuriousTab\n"); In the printf statement, it skips the first 5 characters and it prints "CURIOUSTAB"

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