为什么这个boost :: variant示例不起作用？

Question

我开始了解boost :: variant.我认为这个例子应该有效.

#include <boost/fusion/sequence.hpp>
#include <boost/fusion/include/sequence.hpp>

#include <boost/variant/variant.hpp>
#include <string>
#include <vector>
#include <iostream>
#include <boost/variant/get.hpp>
boost::variant< bool,long,double,std::string,
std::vector<boost::variant<bool> > > v4;
void main()
{

    std::vector<boost::variant<bool> > av (1);
    v4= av;
    try
    {
    bool b=
    boost::get<bool> (v4[0]); // <--- this is line 20
    std::cout << b;


    }
    catch (boost::bad_get v)
    {
    std::cout << "bad get" <<std::endl; 
    }
}

我收到编译错误:

d:\ m\upp\boosttest\main.cpp(20):错误C2676:二进制'[':'boost :: variant'不会定义此运算符或转换为预定义运算符可接受的类型[T0_ = bool,T1 = long,T2 = double,T3 = std :: string,T4 = std :: vector>]

Answer 1

v4[0]因为v4是变体而不是向量,所以无效.您需要先使用它boost::get来检索存储在其中的向量.所以,第20行应该是

boost::get<bool>(boost::get<std::vector<boost::variant<bool> > >(v4)[0]);