关于Lambda与boost::adapters::filtered()一起使用的两个问题

Question

请考虑此非编译代码：

#include <boost/range/adaptors.hpp>
class Stuff {
public:
bool var;
};
class Manager {
/// Get everything
std::vector<Stuff*>
get_all_stuff() const
{
return list_of_stuff;
}
/// Get a vector of only those that whose "var" matches the "tf" argument.
std::vector<Stuff*>
get_some_stuff(const bool tf) const
{
return (get_all_stuff() |
boost::adaptors::filtered(
[](Stuff const& s) { return (s.var == tf); }
)
);
}
private:
std::vector<Stuff*> list_of_stuff;
};

编译死于此错误：

ex.cc: In lambda function:
ex.cc:21:46: error: ‘tf’ is not captured
[](Stuff const& s) { return (s.var == tf); }
^

1.(如何将该函数参数引入lambda？

2.(这是一种危险的做法吗？我应该使用std:：remove_copy_if((吗？

• 我不担心"；get_all_stuff((">
• 我担心"；get_some_stuff((">

Answer 1

要将外部值获取到lambda中，必须捕获它。

[&tf](Stuff const& s) { return (s.var == tf);

我在示例中使用了boost::adaptors::filter。但以太一将返回一个范围，而不是向量对象。如果你想返回一个不同于list_of_stuff的向量，你必须构建它。如果你从函数中返回它，编译器会在可能的情况下移动它。下面是一个关于coliru的工作示例。

#include <iostream>
#include <boost/range/algorithm.hpp>
#include <boost/range/adaptors.hpp>
class Stuff {
public:
bool var;
int id;
};
std::ostream& operator << (std::ostream& os, const Stuff stuff) {
return os << std::boolalpha << stuff.id << " " << stuff.var;
}
using vector_type = std::vector<Stuff>;
class Manager {
/// Get everything
public:
auto get_all_stuff() const
{
return list_of_stuff;
}
// Get a vector of only those that whose "var" matches the "tf" argument.
vector_type get_some_stuff(const bool tf) const
{
vector_type temp;
for (auto item : boost::adaptors::filter(list_of_stuff,
[&tf](Stuff const& s) { return s.var == tf; }))
temp.push_back(item);
return temp;
}
private:
vector_type list_of_stuff = { {false,1},{true,2},{false,3},{true,4},{true,5} };
};
int main()
{
Manager manage;
for (const auto item : manage.get_all_stuff())
std::cout << item << " ";
std::cout << std::endl;
for (const auto item : manage.get_some_stuff(true))
std::cout << item << " ";
std::cout << std::endl;
}