We know in string literal, "\u94b1" will be converted to a character, in this case a Chinese word '钱'. But if it is literally 6 character in a string, saying '\', 'u', '9', '4', 'b', '1', how can I convert it to a character manually?
For example:
string s1;
string s2 = "\u94b1";
cin >> s1; //here I input \u94b1
cout << s1 << endl; //here output \u94b1
cout << s2 << endl; //and here output 钱
I want to convert s1
so that cout << s1 << endl;
will also output 钱
.
Any suggestion please?
In fact the conversion is a little more complicated.
string s2 = "\u94b1";
is in fact the equivalent of:
char cs2 = { 0xe9, 0x92, 0xb1, 0}; string s2 = cs2;
That means that you are initializing it the the 3 characters that compose the UTF8 representation of 钱 - you char just examine s2.c_str()
to make sure of that.
So to process the 6 raw characters '\', 'u', '9', '4', 'b', '1', you must first extract the wchar_t from string s1 = "\\u94b1";
(what you get when you read it). It is easy, just skip the two first characters and read it as hexadecimal:
unsigned int ui;
std::istringstream is(s1.c_str() + 2);
is >> hex >> ui;
ui
is now 0x94b1
.
Now provided you have a C++11 compliant system, you can convert it with std::convert_utf8
:
wchar_t wc = ui;
std::codecvt_utf8<wchar_t> conv;
const wchar_t *wnext;
char *next;
char cbuf[4] = {0}; // initialize the buffer to 0 to have a terminating null
std::mbstate_t state;
conv.out(state, &wc, &wc + 1, wnext, cbuf, cbuf+4, next);
cbuf
contains now the 3 characters representing 钱 in utf8 and a terminating null, and you finaly can do:
string s3 = cbuf;
cout << s3 << endl;
You do this by writing code that checks whether the string contains a backslash, a letter u, and four hexadecimal digits, and converts this to a Unicode code point. Then your std::string implementation probably assumes UTF-8, so you translate that code point into 1, 2, or 3 UTF-8 bytes.
For extra points, figure out how to enter code points outside the basic plane.
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