Hi,
But, when we define any character array as
char a[3]={'a','b'};
then compiler here automatically consider '\0' null character to represent the end of the string

So, my question here is:
As compiler do not needs to use any such terminator to indicate that the int array or float array is terminated. Then, why compiler needs to assign '\0' null terminator in char array & why not in others?

That's not true. You have explicitly initialised two elements of three-element array. The standards go about it in a round-about way (there's a logic train to follow to get to the conclusion, and the C and C++ standards have different logic trains) but the end result is that a[2] in your example will get a value of zero.

You will get the same effect (other than type of elements) with

C Syntax (Toggle Plain Text)

   int a[3] = {1, 2};

int a[3] = {1, 2};

ie a[2] will get the value of zero.




It doesn't. If the number of explicitly initialised elements of an array is less that the size of the array, the standards specify how the other elements are initialised.

There is an anomaly in C (and C++) that allows char arrays to be initialised using strings (and arrays of wchar_t to be initialised with wide strings). Strings, by convention, are zero terminated. There is no equivalent to this style of initialisation for other types. But this anomaly is a different thing from what you're asking about.

So, my query is, why this is happening only in case of strings and not for the others( like int, float, single character)?

There is a difference between a character array and a string. A character array
char a[3] = {'a', 'b', 'c'}; is not a string. Just because it happens to end in a '\0' when you don't define all the elements
char a[3] = {'a', 'b'}; does not make it a string. Don't use it as such.