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Home/ Questions/Q 4266604
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Editorial Team
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Editorial Team
Asked: May 21, 20262026-05-21T06:46:58+00:00 2026-05-21T06:46:58+00:00

This causes problems due to extra whitespace: std::string t(3.14 ); double d = boost::lexical_cast<double>

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This causes problems due to extra whitespace:

std::string t("3.14 ");
double d = boost::lexical_cast<double> (t); 

So, I wrote this

template<typename T> 
T string_convert(const std::string& given)
{
  T output;
  std::stringstream(given) >> output;
  return output;
}

double d = string_convert<double> (t); 

What can be the problems with this? Is there a better way? Much prefer to use lexical cast

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1 Answer

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  1. Editorial Team
    Editorial Team
    2026-05-21T06:46:59+00:00Added an answer on May 21, 2026 at 6:46 am

    Note that your code isn’t always correct. If you do string_convert<double>("a"), for example, the read will fail and you’ll return output uninitialized, leading to undefined behavior.

    You can do this:

    template<typename T> 
    T string_convert(const std::string& given)
    {
      T output;
      if (!(std::stringstream(given) >> output))
        throw std::invalid_argument(); // check that extraction succeeded
    
      return output;
    }
    

    Note the only difference between the above code and Boost’s is that Boost also checks to make sure nothing is left in the stream. What you should do, though, is just trim your string first:

    #include <boost/algorithm/string/trim.hpp>
    
    std::string t("3.14 ");
    boost::algorithm::trim(t); // get rid of surrounding whitespace
    
    double d = boost::lexical_cast<double>(t); 
    
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