Size of escaped characters in C

Why does the following program output 5?

#include <stdio.h> 
main() 
{ 
    char str[]="S\开发者_Python百科065AB"; 
    printf("\n%d", sizeof(str)); 
}

Short answer: See David Heffernan's answer.

Long answer:

§ 6.4.4.4 of the C(99) standard specifies "character constants", which (among others) include simple escape sequences (e.g. '\n', '\\'), octal escape sequences (e.g. '\0'), hexadecimal escape sequences (e.g. '\x0f'), and universal character names (e.g. '\u0112').

The backslash in your example introduces such an escape / octal / hex / universal constant. The following octal digit ([0-7]) makes it an octal constant (hex would be '\x', universal would be '\u', escape sequence would be '\['"?\abfnrtv]').

That octal constant is terminated once three octal digits are consumed, or a non-octal-digit is encountered.

I.e., '\065' is equivalent to '\x35' or (decimal) 53, which is (coincidentally) '5' on the ASCII table - a single character, anyway.

It's the size of the array which has five elements: S, \065, A, B, \0