How many bytes does a string take? A char?

The following will take x consecutive chars in memory:

'n' - 1 char (type char)
"n" - 2 chars (above plus zero character) (type const char[2])
'\n' - 1 char
"\n" - 2 chars
"\\n" - 3 chars ('\', 'n', and zero)
"" - 1 char

edit: formatting fixed

edit2: I've written something very stupid, thanks Mooing Duck for pointing that out.

#include <iostream>
 
int main()
{
    std::cout << sizeof 'n'   << std::endl;   // 1
    std::cout << sizeof "n"   << std::endl;   // 2
    std::cout << sizeof '\n'  << std::endl;   // 1
    std::cout << sizeof "\n"  << std::endl;   // 2
    std::cout << sizeof "\\n" << std::endl;   // 3
    std::cout << sizeof ""    << std::endl;   // 1
}

• Single quotes indicate characters.
• Double quotes indicate C-style strings with an invisible NUL terminator.

\n (line break) is only a single char and so is \\ (backslash). \\n is just a backslash followed by n.

• A char, by definition, takes up one byte.
• Literals using ' are char literals; literals using " are string literals.
• A string literal is implicitly null-terminated, so it will take up one more byte than the observable number of characters in the literal.
• \ is the escape character and \n is a newline character.

Put these together and you should be able to figure it out.

• 'n': is not a string, is a literal char, one byte, the character code for the letter n.
• "n": string, two bytes, one for n and one for the null character every string has at the end.
• "\n": two bytes as \n stand for "new line" which takes one byte, plus one byte for the null char.
• '\n': same as the first, literal char, not a string, one byte.
• "\\n": three bytes.. one for \, one for newline and one for the null character
• "": one byte, just the null character.