What will be the output of the program in Turbo C?

#include<stdio.h>

int main()
{
    char str[10] = "India";
    str[6] = "CURIOUSTAB";
    printf("%s\n", str);
    return 0;
}

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".

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 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.

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".

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: 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

Step 1:

printf("%d, %d, %d", sizeof(3.0f), sizeof('3'), sizeof(3.0));

The sizeof function returns the size of the given expression.

sizeof(3.0f) is a floating point constant. The size of float is 4 bytes

sizeof('3') It converts '3' in to ASCII value.. The size of int is 2 bytes

sizeof(3.0) is a double constant. The size of double is 8 bytes

Hence the output of the program is 4,2,8

Note: The above program may produce different output in other platform due to the platform dependency of C compiler.

In Turbo C, 4 2 8. But in GCC, the output will be 4 4 8.

Here str[] has declared as 7 character array and into a 8 character is stored. This will result in overwriting of the byte beyond 7 byte reserved for '\0'.