lexical_cast的代码段：
class lexical_castable {
public:
lexical_castable() {};
lexical_castable(const std::string s) : s_(s) {};
friend std::ostream operator<<
(std::ostream& o,const lexical_castable& le);
friend std::istream operator>>
(std::istream& i,lexical_castable& le);
private:
virtual void print_(std::ostream& o) const {
o << s_ <<"\n";
}
virtual void read_(std::istream& i) const {
i >> s_;
}
std::string s_;
};
std::ostream operator<<(std::ostream& o,const lexical_castable& le) {
le.print_(o);
return o;
}
std::istream operator>>(std::istream& i,lexical_castable& le) {
le.read_(i);
return i;
}
基于document,
template<typename Target,typename Source> Target lexical_cast(const Source& arg);
1> Returns the result of streaming arg into a standard library
string-based stream and then out as a Target object.2> Source is OutputStreamable
3> Target is InputStreamable
问题1>对于用户定义类型(UDT),OutputStreamable或InputStreamable是否总是必须处理std :: string？例如,给定一个包含一个简单整数作为成员变量的类,当我们定义运算符<<和运算符>>,实现代码是什么样的？我必须将整数转换为字符串吗？根据我的理解,似乎UDT总是必须处理std :: string才能使用boost :: lexical_cast和boost :: lexcial_cast需要中间std :: string来进行真正的转换作业.
问题2>为什么运算符的返回值<<或运算符>>在上面的代码中没有引用std :: ostream&或std :: istream&分别？
解决方法
要使您的类可用于lexical_cast,只需为其定义“stream”运算符即可.
从 Boost.LexicalCast Synopsis开始：
从 Boost.LexicalCast Synopsis开始：
- Source is OutputStreamable,meaning that an
operator<<is defined that takes astd::ostreamorstd::wostreamobject on the left hand side and an instance of the argument type on the right.- Target is InputStreamable,meaning that an
operator>>is defined that takes astd::istreamorstd::wistreamobject on the left hand side and an instance of the result type on the right.- Target is copyConstructible [20.1.3].
- Target is DefaultConstructible,meaning that it is possible to default-initialize an object of that type [8.5,20.1.4].
：
// either inline friend,out-of-class friend,or just normal free function
// depending on whether it needs to access internel members
// or can cope with the public interface
// (use only one version)
class MyClass{
int _i;
public:
// inline version
friend std::ostream& operator<<(std::ostream& os,MyClass const& ms){
return os << ms._i;
}
// or out-of-class friend (friend declaration inside class only)
friend std::ostream& operator<<(std::ostream& os,MyClass const& ms);
// for the free function version
int get_i() const{ return _i; }
};
// out-of-class continued
std::ostream& operator<<(std::ostream& os,MyClass const& ms){
return os << ms._i;
}
// free function,non-friend
std::ostream& operator<<(std::ostream& os,MyClass const& ms){
return os << ms.get_i();
}
对于操作符>>当然也是如此.