当我们使用cout <时，boost :: any和boost :: variant之间的区别

Question

I have run the following piece of code successfully using boost::variant<string, float> . However, when I tried to use boost::any instead, I face mismatch operand error at cout<< instruction, the piece of code is :

for( vector<vector<vector<boost::any>>>::const_iterator i = masterList.begin(); i != masterList.end(); ++i)
{
    for( vector<vector<boost::any>>::const_iterator j = i->begin(); j != i->end(); ++j)
    {
        for( vector<boost::any>::const_iterator k = j->begin(); k != j->end(); ++k)
        { 
            cout<<*k<<' ';
        }
    }
}

Answer 1

Boost.Any offers full type erasure, all characteristics (such as streaming to a std::ostream ) of the underlying type are erased. The only way to get back the type is by using the any_cast functions.

If you want partial type erasure, look at the Boost.TypeErasure library of Steven Watanabe. Note that TypeErasure is an official Boost library since Boost 1.54.