C++: size of a char array using sizeof

Question

Look at the following piece of code in C++:

char a1[] = {'a','b','c'};
char a2[] = "abc";
cout << sizeof(a1) << endl << sizeof(a2) << endl;

Though sizeof(char) is 1 byte, why does the output show sizeof(a2) as 4 and not 3 (as in case of a1 )?

Answer 1

C-strings contain a null terminator, thus adding a character.

Essentially this:

char a2[] = {'a','b','c','\0'};

Answer 2

That's because there's an extra null '\\0' character added to the end of the C-string, whereas the first variable, a1 is an array of three seperate characters.

sizeof will tell you the byte size of a variable, but prefer strlen if you want the length of a C-string at runtime.

Answer 3

For a2, this is a string so it also contains the '\\n'

Correction, after Ethan & Adam comment, this is not '\\n' of course but null terminator which is '\\0'