what will be output of following program ?

Question

#include

main()

{

   char str[]="S¥065AB";

printf("¥\n%d", sizeof(str));

}

Answer 1

the output is :
¥

9

the question is why?
because this character "¥", is presented by two bytes (UTF8)

Answer 2

Try if yourself

First of all, you can just run it and see for yourself.

Second, if you try to run it, you'll see a compilation error. You're including something, but not telling the compiler what it is. (It's most probably supposed to be stdio.h.

If you fix the include, the program outputs the ¥ sign and the size of the str variable (character array).

Actual result

Now, what the actual result is?

To kill the suspense, the program prints 9.

Why?

Well, the array itself contains elements of a character sequence. The character sequence (I am deliberately not calling it string) contains 7 characters (S¥065AB). However, the array needs to store the terminating zero for the string, so there's the '\0' character as well. So that's now 7 + 1 = 8.

The next important thing is that the ¥ sign seems like to be a Unicode character, so it takes up 2 bytes, not one. Hence 6 bytes (1 byte for each ASCII characters) + 2 bytes (for the single Unicode character) + 1 byte (terminating zero) = 9 bytes

Answer 3

with this sign we get :¥

9