C++: size of a char array using sizeof

问题

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)?

回答1:

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

Essentially this:

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

回答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.

回答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'

来源：https://stackoverflow.com/questions/10735990/c-size-of-a-char-array-using-sizeof