How to convert std::wstring to numeric type(int, l

问题:

What's the best way to convert std::wstring to numeric type, such as int, long, float or double?

回答1:

Either use boost::lexical_cast<>:

#include <boost/lexical_cast.hpp>

std::wstring s1(L"123");
int num = boost::lexical_cast<int>(s1);

std::wstring s2(L"123.5");
double d = boost::lexical_cast<double>(s2);

These will throw a boost::bad_lexical_cast exception if the string can't be converted.

The other option is to use Boost Qi (a sublibrary of Boost.Spirit):

#include <boost/spirit/include/qi.hpp>

std::wstring s1(L"123");
int num = 0;
if (boost::spirit::qi::parse(s1.begin(), s1.end(), num))
    ; // conversion successful

std::wstring s2(L"123.5");
double d = 0;
if (boost::spirit::qi::parse(s1.begin(), s1.end(), d))
    ; // conversion successful

Using Qi is much faster than lexical_cast but will increase your compile times.

回答2:

C++0x introduces the following functions in <string>:

int                stoi  (const wstring& str, size_t* idx = 0, int base = 10);
long               stol  (const wstring& str, size_t* idx = 0, int base = 10);
unsigned long      stoul (const wstring& str, size_t* idx = 0, int base = 10);
long long          stoll (const wstring& str, size_t* idx = 0, int base = 10);
unsigned long long stoull(const wstring& str, size_t* idx = 0, int base = 10);

float       stof (const wstring& str, size_t* idx = 0);
double      stod (const wstring& str, size_t* idx = 0);
long double stold(const wstring& str, size_t* idx = 0);

idx is an optionally null pointer to the end of the conversion within str (set by the conversion function).

回答3:

Best?

If you don't want to use anything more than the CRT library, and are happy with getting 0 if the string cannot be converted, then you can save on error handling, complex syntax, including headers by

std::wstring s(L"123.5");
float value = (float) _wtof( s.c_str() );

It all depends what you are doing. This is the KISS way!

回答4:

Use wstringstream / stringstream:

#include <sstream>
float toFloat(const std::wstring& strbuf)
{
    std::wstringstream converter;
    float value = 0;

    converter.precision(4);
    converter.fill('0');
    converter.setf( std::ios::fixed, std::ios::floatfield );                              

    converter << strbuf;
    converter >> value;
    return value;
}

回答5:

just use the stringstream: do not forget to #include <sstream>

wchar_t blank;
wstring sInt(L"123");
wstring sFloat(L"123.456");
wstring sLong(L"1234567890");
int rInt;
float rFloat;
long rLong;

wstringstream convStream;

convStream << sInt<<' '<< sFloat<<' '<<sLong;
convStream >> rInt;
convStream >> rFloat;
convStream >> rLong;
cout << rInt << endl << rFloat << endl << rLong << endl;

回答6:

So I was using Embarcadero and that piece of ..... didn´t let me use stoi, so i have to create my own function.

int string_int(wstring lala){
    int numero;
    int completo = 0;
    int exponente = 1;
    int l = 1;
    for (int i = 0; i < lala.size(); i++){
        numero = 0;
        for (int j = 48; j < 57; j++){
            if (lala[i] == j){
                break;
            }
            numero++;
        }
        for (int k = 0; k < lala.size() - l; k++){
            exponente *= 10;
        }
        numero *= exponente;
        completo += numero;
        exponente = 1;
        l++;
    }
    return completo;
}