fusion::vector + fusion::push_back = fusion::vector?

Question

I typedef a vector with two elements. Then I push_back into it an other element and expect what result type is also a vector. But that's not so.

Example:

typedef boost::fusion::vector<int, double> vec1;
typedef boost::fusion::result_of::push_back<vec1, std::string> vec2;
//boost::is_same<vec2, boost::fusion::vector<int, double, std::string>>::value == false

http://liveworkspace.org/code/361492801eebe24cc5679a1e899a5240

What am I doing wrong?

Regards.

Cat Plus Plus · Accepted Answer · 2011-12-03T10:08:10.667

2

You've aliased the push_back itself as vec2. You need to use

typedef boost::fusion::result_of::push_back<vec1, std::string>::type vec2;

But keep in mind that the type might still not be the same, Fusion algorithms are not required to preserve the type (and since push_back function is supposed to return a lazy view, then vec2 will most likely be some view type). The only guarantee is that vec2 will be "A model of Forward Sequence.".

edited Dec 03 '11 at 10:08

answered Dec 03 '11 at 10:02

Cat Plus Plus

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Please tell me, how can I initialize the object of type vec2? – niXman Dec 03 '11 at 10:18
vec2 v(33, .14, "string"); link: http://liveworkspace.org/code/6d35d3211640ee62174dced31a28d937 – niXman Dec 03 '11 at 10:22
@niXman: `vec2 v(boost::fusion::push_back(vec1(33, .14), "string"));` I guess. I don't really use Fusion. – Cat Plus Plus Dec 03 '11 at 10:25
Thanks. I`ve expected a more usability. – niXman Dec 03 '11 at 10:29
1

You can use as_vector to force the return type of the push_back to act as a vector properly – Joel Falcou Dec 03 '11 at 14:07

fusion::vector + fusion::push_back = fusion::vector?

1 Answers1