What will be the output of the program?

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

int main()
{
    static char str1[] = "dills";
    static char str2[20];
    static char str3[] = "Daffo";
    int i;
    i = strcmp(strcat(str3, strcpy(str2, str1)), "Daffodills");
    printf("%d\n", i);
    return 0;
}

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

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+"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'.

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".

The function strlen returns the number of characters in the given string.

Therefore, strlen("123456") returns 6.

Hence the output of the program is "6".