how can we write a wrapper lexical cast function in order to implement lines like :
int value = lexical_cast<int> (string) I am quite new to programming and was wondering how we can write the function. I don't know how to figure out a template. Also can we write a wrapper function for double too ? Like
double value = lexical_cast2<double> (string) ??
To have it as you stated in your example:
#include <sstream> template <class Dest> class lexical_cast { Dest value; public: template <class Src> lexical_cast(const Src &src) { std::stringstream s; s << src; s >> value; } operator const Dest &() const { return value; } operator Dest &() { return value; } }; Including error checking:
template <class Src> lexical_cast(const Src &src) throw (const char*) { std::stringstream s; if (!(s << src) || !(s >> value) || s.rdbuf()->in_avail()) { throw "value error"; } } 
you could try something like this:
#include <sstream> #include <iostream> template <class T> void FromString ( T & t, const std::string &s ) { std::stringstream str; str << s; str >> t; } int main() { std::string myString("42.0"); double value = 0.0; FromString(value,myString); std::cout << "The answer to all questions: " << value; return 0; } 
double value = FromString<double>(myString);
FromString<double>(value,myString);If this is not an excersice and if your goal is just to convert string to other types:
If you are using C++11, there are new conversion functions.
so you can do something like
std::stoi -> int std::stol -> long int std::stoul -> unsigned int std::stoll -> long long std::stoull -> unsigned long long std::stof -> float std::stod -> double std::stold -> long double http://www.cplusplus.com/reference/string/
If not C++11 you can use
int i = atoi( my_string.c_str() ) double l = atof( my_string.c_str() ); 