What will be the output of the program?

#include<stdio.h>

int main()
{
    char str[] = "Nagpur";
    str[0]='K';
    printf("%s, ", str);
    str = "Kanpur";
    printf("%s", str+1);
    return 0;
}

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".

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

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

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.

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'.

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

Hence the output is "Morning"

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"