Here's an Ideone: http://ideone.com/vjByty.
#include <iostream> using namespace std; #include <string> int main() { string s = "\u0001\u0001"; cout << s.length() << endl; if (s[0] == s[1]) { cout << "equal\n"; } return 0; } I'm confused on so many levels.
What does it mean when I type in an escaped Unicode string literal in my C++ program?
Shouldn't it take 4 bytes for 2 characters? (assuming utf-16)
Why are the first two characters of s (first two bytes) equal?
So the draft C++11 standard says the following about universal characters in narrow string literals (emphasis mine going forward):
Escape sequences and universal-character-names in non-raw string literals have the same meaning as in character literals (2.14.3), except that the single quote [...] In a narrow string literal, a universal-charactername may map to more than one char element due to multibyte encoding
and includes the following note:
The size of a narrow string literal is the total number of escape sequences and other characters, plus at least one for the multibyte encoding of each universal-character-name, plus one for the terminating ’\0’.
Section 2.14.3 referred to above says:
A universal-character-name is translated to the encoding, in the appropriate execution character set, of the character named. If there is no such encoding, the universal-character-name is translated to an implementation defined encoding.
if I try this example (see it live):
string s = "\u0F01\u0001"; The first universal character does map to multiple characters.
What does it mean when I type in an escaped Unicode string literal in my C++ program?
To quote the standard:
A universal-character-name is translated to the encoding, in the appropriate execution character set, of the character named. If there is no such encoding, the universal-character-name is translated to an implementation-defined encoding.
Typically, the execution character set will be ASCII, which contains a character with value 1. So \u0001 will be translated into a single character with value 1.
If you were to specify non-ASCII characters, like \u263A, you might see more than one byte per character.
Shouldn't it take 4 bytes for 2 characters? (assuming utf-16)
If it were UTF-16, yes. But string can't be encoded with UTF-16, unless char has 16 bits, which it usually doesn't. UTF-8 is a more likely encoding, in which characters with values up to 127 (that is, the whole ASCII set) are encoded with a single byte.
Why are the first two characters of s (first two bytes) equal?
With the above assumptions, they are both the character with value 1.
'\u0001' will translate into a single byte.
\u0001 must translate to a single byte, as the answer says. Perhaps "more likely" was a poor choice of words, since apparently there are some quaint systems that still use 8-bit encodings. But I've no idea what they might do with arbitrary Unicode points, and don't really want to know.
\u0001\u0000. ideone.com/N6KkHt Both the characters are still treated as identical. They should both be treated as ascii and that means they should be treated as 1 and 0 respectively.\u0000 is incorrectly translated to 1. gcc.gnu.org/bugzilla/show_bug.cgi?id=53690
Lon the string literal, then I suppose it's often. But that's not what we have here.char, with the UTF-16 as the basic execution character set. (I don't know of any that do, but the standard definitely allows it.)