我在使用 boost::lexical_cast 时遇到问题。我正在尝试在 GLM(OpenGL 数学)库中的一个类上使用它。
为了允许词法转换,我已经为相关类实现了 operator<< 函数:
template <class T>
std::ostream& operator<<(std::ostream& out, const glm::detail::tvec2<T>& vec)
{
out << vec.x << " " << vec.y;
return out;
}
template <class T>
std::istream& operator>>(std::istream& in, glm::detail::tvec2<T>& vec)
{
in >> vec.x;
in >> vec.y;
return in;
}
我像这样测试了运算符:
std::cout << glm::ivec2(1, 1) << glm::vec2(1.0f, 1.0f);
和:
std::stringstream ss("640 480");
glm::ivec2 pt;
ss >> pt;
std::cout << pt << std::endl;
这很好用,但是如果我尝试这样做:
glm::ivec2 pt = boost::lexical_cast<glm::ivec2>("1 1");
我收到以下错误:
/usr/include/boost/lexical_cast.hpp: In member function ‘bool boost::detail::lexical_stream_limited_src<CharT, Base, Traits>::operator>>(InputStreamable&) [with InputStreamable = glm::detail::tvec2<int>, CharT = char, Base = std::basic_streambuf<char>, Traits = std::char_traits<char>]’:
/usr/include/boost/lexical_cast.hpp:1151:13: instantiated from ‘Target boost::detail::lexical_cast(typename boost::call_traits<B>::param_type, CharT*, std::size_t) [with Target = glm::detail::tvec2<int>, Source = const char*, bool Unlimited = false, CharT = char, typename boost::call_traits<B>::param_type = const char* const, std::size_t = long unsigned int]’
/usr/include/boost/lexical_cast.hpp:1174:77: instantiated from ‘Target boost::lexical_cast(const Source&) [with Target = glm::detail::tvec2<int>, Source = char [8]]’
test2.cpp:41:59: instantiated from here
/usr/include/boost/lexical_cast.hpp:785:29: error: cannot bind ‘std::basic_istream<char>’ lvalue to ‘std::basic_istream<char>&&’
/usr/include/c++/4.6/istream:852:5: error: initializing argument 1 of ‘std::basic_istream<_CharT, _Traits>& std::operator>>(std::basic_istream<_CharT, _Traits>&&, _Tp&) [with _CharT = char, _Traits = std::char_traits<char>, _Tp = glm::detail::tvec2<int>]’
编辑:似乎只有包含 Boost.PropertyTree 的 header 时才会出现错误。
最佳答案
您发布了这个提取运算符:
template <class T>
std::istream& operator>>(std::istream& in, glm::detail::tvec2<T>& vec)
但错误暗示它正在尝试用这个编译:
std::istream& std::operator>>(std::istream&&, _Tp&)
[with ... _Tp = glm::detail::tvec2<int>]
您确定您发布的代码与您正在编译的代码匹配吗? (如果您不能立即看到它,请关注运算符的第一个参数)。
好的,这个完整的代码对我来说很好用:
#include <sstream>
#include <iostream>
#include <boost/lexical_cast.hpp>
template <class T>
struct tvec2 {
tvec2() : x(), y() {}
tvec2(tvec2 const &) = default;
tvec2(T x_, T y_) : x(x_), y(y_) {}
T x;
T y;
};
template <class T>
std::ostream& operator<<(std::ostream& out, const tvec2<T>& vec)
{
out << vec.x << " " << vec.y;
return out;
}
template <class T>
std::istream& operator>>(std::istream& in, tvec2<T>& vec)
{
// yuck, boost disables skipws on the input stream
in >> vec.x;
if (in.good() && in.ignore(256, ' ').good())
in >> vec.y;
return in;
}
void test_operators()
{
tvec2<int> intvec(2,3);
std::cout << "intvec = {" << intvec << "}\n";
std::stringstream ss;
ss << intvec;
tvec2<int> dupvec;
ss >> dupvec;
std::cout << "dupvec = {" << dupvec << "}\n";
}
void test_lexical_cast()
{
std::cout << "and now with lexical_cast ...\n";
tvec2<int> dupvec = boost::lexical_cast<tvec2<int> >("2 3");
std::cout << "dupvec = {" << dupvec << "}\n";
}
int main()
{
test_operators();
test_lexical_cast();
}
关于c++ - 我如何让 Boost.LexicalCast 工作?,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/7904423/