Why does the size of this std::string change, when characters are changed?

Question

I have an issue in which the size of the string is effected with the presence of a '\0' character. I searched all over in SO and could not get the answer still.

Here is the snippet.

int main() {   std::string a = "123123\0shai\0";   std::cout << a.length(); } 

http://ideone.com/W6Bhfl

The output in this case is

6 

Where as the same program with a different string having numerals instead of characters

int main() {   std::string a = "123123\0123\0";   std::cout << a.length(); } 

http://ideone.com/mtfS50

gives an output of

8 

What exactly is happening under the hood? How does presence of a '\0' character change the behavior?

Answer 1

The sequence \012 when used in a string (or character) literal is an octal escape sequence. It's the octal number 12 which corresponds to the ASCII linefeed ('\n') character.

That means your second string is actually equal to "123123\n3\0" (plus the actual string literal terminator).

It would have been very clear if you tried to print the contents of the string.

Octal sequences are one to three digits long, and the compiler will use as many digits as possible.

Answer 2

If you check the coloring at ideone you will see that \012 has a different color. That is because this is a single character written in octal.