如何将std :: wstring转换为数字类型（int，long，float）？ [英] How to convert std::wstring to numeric type(int, long, float)?

问题描述

将std :: wstring转换为数字类型（例如int，long，float或double）的最佳方法是什么？

解决方案

p>使用 boost :: lexical_cast<> ：

  #include< boost / lexical_cast.hpp> 
 
 std :: wstring s1（L123）; 
 int num = boost :: lexical_cast< int>（s1）; 
 
 std :: wstring s2（L123.5）; 
 double d = boost :: lexical_cast< double>（s2）; 
  

这些会抛出一个 boost :: bad_lexical_cast

另一个选项是使用Boost Qi（Boost.Spirit的子库）：

  #include< boost / spirit / include / qi.hpp> 
 
 std :: wstring s1（L123）; 
 int num = 0; 
 if（boost :: spirit :: qi :: parse（s1.begin（），s1.end（），num））
; // conversion successful 
 
 std :: wstring s2（L123.5）; 
 double d = 0; 
 if（boost :: spirit :: qi :: parse（s1.begin（），s1.end（），d））
; // conversion successful 
  

使用Qi比lexical_cast更快，但会增加你的编译时间。 >

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

解决方案

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.

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