C ++文字和Unicode

Question

C++ Literals

Environment:

• OS: Windows 10 Pro;
• Compiler: GCC latest.
• IDE: Code::Blocks latest.
• working on: Console applications.

My understanding for numerical literals prefixes is that they are useful to determine the numerical value type (not sure).However, I have a lot of confusion on character and string literals prefixes and suffixes. I read a lot and spent days trying to understand the situation, but I got more questions and few answers. so I thought stack overflow could be of a lot of help.

Qs:

1- What are the correct use for the string prefixes u8 u UL?

I have the following code as example:

#include <iostream>
#include <string>
using namespace std;

int main()
{
    cout << "\n\n Hello World! (plain) \n";
    cout << u8"\n Hello World! (u8) \n";
    cout << u"\n Hello World! (u) \n";
    cout << U"\n Hello World! (U) \n";
    cout << L"\n Hello World! (plain) \n\n";

    cout << "\n\n\n";
}

The output is like this:

Hello World! (plain)

Hello World! (u8)

0x47f0580x47f0840x47f0d8

Q2: Why U u ans L has such output? I expected it is just to determine type not do encoding mapping (if it is).

Q3 Is there a simple and to the point references about encodings like UTF-8. I am confused about them, in addition I doubt that console applications is capable to deal with them. I see it is crucial to understand them.

Q4: Also I will appreciate a step by step reference that explain custom type literals.

Answer 1

First see: http://en.cppreference.com/w/cpp/language/string_literal

std::cout 's class operator << is properly overloaded to print const char* . That is why the first two strings are printed.

 cout << "\\n\\n Hello World! (plain) \\n"; cout << u8"\\n Hello World! (u8) \\n"; 

As expected, prints ¹ :

 Hello World! (plain) Hello World! (u8) 

Meanwhile std::cout 's class has no special << overload for const char16_t* , const char32_t* and const wchar_t* , hence it will match << 's overload for printing pointers, that is why:

 cout << u"\\n Hello World! (u) \\n"; cout << U"\\n Hello World! (U) \\n"; cout << L"\\n Hello World! (plain) \\n\\n"; 

Prints:

 0x47f0580x47f0840x47f0d8 

As you can see, there are actually 3 pointer values printed there: 0x47f058 , 0x47f084 and 0x47f0d8

---

However, for the last one, you can get it to print properly using std::wcout

 std::wcout << L"\\n Hello World! (plain) \\n\\n"; 

prints

  Hello World! (plain) 

---

^{1: The u8 literal printed as expected because of the direct ASCII mapping of the first few codepoints of UTF-8.}

Answer 2

1) Narrow multibyte string literal. The type of an unprefixed string literal is const char[] .

2) Wide string literal. The type of a L"..." string literal is const wchar_t[] .

3) UTF-8 encoded string literal. The type of a u8"..." string literal is const char[] .

4) UTF-16 encoded string literal. The type of a u"..." string literal is const char16_t[] .

5) UTF-32 encoded string literal. The type of a U"..." string literal is const char32_t[] .

6) Raw string literal. Used to avoid escaping of any character, anything between the delimiters becomes part of the string. prefix, if present, has the same meaning as described above.

std::cout expects single byte characters, otherwise it can output a value such as 0x47f0580x47f0840x47f0d8 . If your trying to output literals that consist of multi-byte characters (char16_t, char32_t, or wchar_t) then you need to use std::wcout to output them to the console, or convert them to a single byte character type. Raw string literals are very handy for formatting output. An example of Raw string literals is R"~(This is the text that will be output just as I typed it into the code editor!)~" and will be a single byte character string. If it's prefixed with any of the multi-byte qualifiers the raw string literal will be multi-byte. Here is a very comprehensive reference on string literals.