boost::向量中元素的变体和打印方法
std::vector< boost::variant<std::string, int> > vec;
std::string s1("abacus");
int i1 = 42;
vec.push_back(s1);
vec.push_back(i1);
std::cout << vec.at(0).size() << "\n";
当我尝试运行此代码时，出现以下错误：
main.cpp:68: error: ‘class boost::variant<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, int, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_>’ has no member named ‘size’
make: *** [main.o] Error 1
但是，作为字符串，它应该有一个 size() 方法。我不确定出了什么问题。请注意，将最后一行替换为：
std::cout << vec.at(0) << "\n";
将按预期打印“abacus”。
std::vector< boost::variant<std::string, int> > vec;
std::string s1("abacus");
int i1 = 42;
vec.push_back(s1);
vec.push_back(i1);
std::cout << vec.at(0).size() << "\n";
when I try to run this code, I get the following error:
main.cpp:68: error: ‘class boost::variant<std::basic_string<char, std::char_traits<char>, std::allocator<char> >, int, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_, boost::detail::variant::void_>’ has no member named ‘size’
make: *** [main.o] Error 1
however, being a string it should have a size() method. I'm not sure what is going wrong. note that replacing the last line with:
std::cout << vec.at(0) << "\n";
will print "abacus", as expected.
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它不是一个
字符串——它是一个变体。您首先需要告诉编译器您知道里面有一个字符串- 即使用boost::get(vec [0]） 。请务必阅读 Boost。变体教程。
It’s not a
string– it’s avariant. You first need to tell the compiler that you know there’s astringinside – i.e. retrieve it usingboost::get<std::string>(vec[0]).Be sure to read the Boost.Variant tutorial.
您需要获取此变体的第一种类型（即字符串），您使用
vector::at()访问的类boost::variant没有名为size()的方法，请尝试以下操作：:You need to get the first type of this variant (which is the string), the class
boost::variantwhich you are accessing withvector::at()has no method calledsize(), try something like::