在C ++ 03中,当您使用运算符typeid时,会返回type_info个对象。
是否可以仅根据此结果检索给定类型的大小,例如由sizeof运算符返回?
例如:
std::type_info info = typeid(int);
int intSize = sizeof(int);
int intSize2 = info.getSize(); // doesn't exist!
问题是我们使用第三方多数组类来返回type_info,但不是类型的大小。
答案 0 :(得分:1)
我能看到的最佳方式(我希望被证明是错误的)是预先注册类型,如下所示:
#include <typeinfo>
#include <iostream>
#include <stdexcept>
#include <map>
#include <vector>
typedef std::map<const std::type_info*, std::size_t> sizes_container; // we cannot use std::type_index, but that's ok - typeid returns const std::type_info&, which refers to an object which lives during the entire lifetime of the program
sizes_container& sizes() // wrapped in a function to avoid static initialization order fiasco
{
static sizes_container s;
return s;
}
template<typename T>
void register_size() // Register the type. Can be called multiple times for the same type.
{
sizes().insert(sizes_container::value_type(&typeid(T), sizeof(T)));
}
class no_such_type : public std::domain_error
{
public:
no_such_type(const std::string& str) :
std::domain_error(str)
{
}
};
std::size_t get_size(const std::type_info& typeinfo)
{
sizes_container::iterator it = sizes().find(&typeinfo);
if(it != sizes().end())
{
return it->second;
}
else
{
throw no_such_type(std::string("No type ") + typeinfo.name() + " registered");
}
}
int main()
{
register_size<int>();
register_size<std::vector<int> >();
std::cout << get_size(typeid(int)) << "\n" // get the size from std::type_info, possibly at runtime
<< get_size(typeid(std::vector<int>)) << "\n" << std::flush;
std::cout << get_size(typeid(long)); // if the type isn't registered, the exception no_such_type is thrown
}
可能的输出:
4
24
This application has requested the Runtime to terminate it in an unusual way.
Please contact the application's support team for more information.
terminate called after throwing an instance of 'no_such_type'
what(): No type l registered
如果您可以控制如何创建数组(例如,使用工厂方法),则可以在此处直接注册类型。