What will be the output of the program?

#include<stdio.h>

int main()
{
    char str1[] = "Hello";
    char str2[10];
    char *t, *s;
    s = str1;
    t = str2;
    while(*t=*s)
        *t++ = *s++;
    printf("%s\n", str2);
    return 0;
}

printf(5+"Good Morning\n"); It skips the 5 characters and prints the given string.

Hence the output is "Morning"

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 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

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".