使用额外的参数提升变体访客 [英] Boost variant visitor with an extra parameter
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问题描述
我的代码与下面类似.
typedef uint32_t IntType;
typedef IntType IntValue;
typedef boost::variant<IntValue, std::string> MsgValue;
MsgValue v;
与其说这个,
IntValue value = boost::apply_visitor(d_string_int_visitor(), v);
我想传递一个这样的额外参数:但是operator()给出了编译错误.
I would like to pass an extra parameter like this: But operator() gives a compile error.
//This gives an error since the overload below doesn't work.
IntValue value = boost::apply_visitor(d_string_int_visitor(), v, anotherStr);
class d_string_int_visitor : public boost::static_visitor<IntType>
{
public:
inline IntType operator()(IntType i) const
{
return i;
}
inline IntValue operator()(const std::string& str) const noexcept
{
// code in here
}
//I want this, but compiler error.
inline IntValue operator()(const std::string& str, const std::string s) const noexcept
{
// code in here
}
};
推荐答案
You can bind the extra string
argument to the visitor using std::bind
. First, add the std::string
parameter to all of the visitor's operator()
overloads.
class d_string_int_visitor : public boost::static_visitor<IntType>
{
public:
inline IntType operator()(IntType i, const std::string& s) const
{
return i;
}
inline IntValue operator()(const std::string& str, const std::string& s) const noexcept
{
// code in here
return 0;
}
};
现在创建一个您已经将第二个 string
参数绑定到的访问者.
Now create a visitor to which you have bound the second string
argument.
auto bound_visitor = std::bind(d_string_int_visitor(), std::placeholders::_1, "Hello World!");
boost::apply_visitor(bound_visitor, v);
但是,更好的解决方案是将字符串作为访问者的构造函数参数传递.
However, a better solution would be to pass the string as the visitor's constructor argument.
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