What's the best way to convert std::wstring to numeric type, such as int, long, float or double?
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possible duplicate of How do you convert a C++ string to an int?Loki Astari– Loki Astari2011-02-25 15:47:26 +00:00Commented Feb 25, 2011 at 15:47
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6 Answers
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).
Comments
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.
1 Comment
Martin Beckett
And people claim C++ is unecessarily complex!
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!
3 Comments
David Thornley
Where does
_wtof come from?
ravenspoint
@david The c run time (CRT) library
David Thornley
I'm not seeing a reference to it either in the C99 Standard or Harbison & Steele. Perhaps wcstod/wcstol/wcstoul would be better.
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;
}
Comments
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;
Comments
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;
}
