Copy members to std::vector

Question

struct Node
{
    std::string name;
    ...
};

typedef std::vector<Node> Nodes;

Nodes nodes;
std::vector<std::string> names;

Is there a nice one-liner way of populating the vector names with Node::name for each item in nodes?

This is what I currently do:

names.reserve(nodes.size());
BOOST_FOREACH(Node node, nodes) 
{
    names.push_back(node.name);
}

I'm using C++98, std and boost.

Could you clarify what you mean by "populating names with Node::name"? — juanchopanza, Oct 24 '13 at 12:34
About the only way you'll get this into a one liner with C++98 is if you make an implicit conversion between `std::string` and `Node` (which is not recommended). — Yuushi, Oct 24 '13 at 12:38
@Blacktempel: "populating the vector names with Node::name" suggests the OP wants to push back only the node names. — thokra, Oct 24 '13 at 12:43
@Yuushi: Thats what I meant with `get the struct into the vector`. Didn't say get the whole struct into names[0], that would be kind of strange. — Blacktempel, Oct 24 '13 at 12:49

David Rodríguez - dribeas · Accepted Answer · 2013-10-24T15:24:34.060

6

This is much simpler with newer libraries (boost) and or standards (C++11), but you should be able to write a small functor:

struct ReadNodeName {
   std::string const & operator()(Node const & node) const {
      return node.name;
   }
};

And then use std::transform

std::transform(nodes.begin(), nodes.end(),
               std::back_inserter(names),
               ReadNodeName());

With boost, the functor could be just boost::bind, in C++11 you could use a lambda: [](Node const & node) { return node.name; };, in C++14 a shorter lambda [](auto& n) { return n.name; }:

std::transform(nodes.begin(), nodes.end(),
               std::back_inserter(names),
               [](auto & n){return n.name;});

Using boost bind -- beware, untested, I don't have access to boost

std::transform(nodes.begin(), nodes.end(),
               std::back_inserter(names),
               boost::bind(&Node::name, _1));

edited Oct 24 '13 at 15:24

answered Oct 24 '13 at 12:45

David Rodríguez - dribeas

204,818
23
294
489

@David Rodríguez - dribeas Do you mind showing us how the boost::bind might work? Thank! – Baz Oct 24 '13 at 12:48
I guess a plain non-member function would also be a good alternative here. – juanchopanza Oct 24 '13 at 12:49
1

@Baz: Added *untested* code, but that should be close to the right thing – David Rodríguez - dribeas Oct 24 '13 at 15:30

score 4 · Answer 2 · answered Oct 24 '13 at 12:38

4

Use std::transform. This takes care of the actual copying in one line. You will have to write a transformation function, though.

answered Oct 24 '13 at 12:38

Oswald

31,254
3
43
68

score 4 · Answer 3 · answered Oct 24 '13 at 12:48

4

If you add a getName() function to Node, you can do this:

std::transform(nodes.begin(), nodes.end(), back_inserter(names), mem_fun(&Node::getName));

answered Oct 24 '13 at 12:48

Tobias Brandt

3,393
18
28

I'd like Node to feel more like a data container than a class. It starts feeling more and more like a class if I add more and more functions to it. This is why I define it using struct instead of class. – Baz Oct 24 '13 at 12:50
@Baz while I understand where you're coming from, adding member functions does not change anything about your object under the hood. It's still plain old data until you start adding in virtual functions etc. – bstamour Oct 24 '13 at 12:59
You could just write a free function that takes a `Node` and returns its `string`, and pass the name of that function as the last argument to transform. – Aaron McDaid Oct 24 '13 at 15:31

Arpegius · Answer 4 · 2013-10-24T13:28:06.730

2

Since you mention boost, here is one liner with boost::lambda:

std::transform(nodes.begin(),nodes.end(), std::back_inserter(names), &boost::lambda::_1 ->* &Node::name );

Boost lambda cannot overload .* so first we need to get address &, and then use ->* to get reference to member.

edited Oct 24 '13 at 13:28

answered Oct 24 '13 at 13:08

Arpegius

5,817
38
53

Copy members to std::vector

4 Answers4