带有数组的C ++异构映射

Question

I have seen people suggest using boost::any to have a heterogeneous map. Ie

map<string,boost::any> 

Has anyone had any luck storing (and successfully using ) arrays stored in the map doing said technique?

I have been able to do the following:

boost::array<char,100> myArr;
map<string,boost::any> myMap;
myArr[51] = a; 
myMap["imageData"] = myArr;

The problem with this is that this seems to store the entire array in memory twice. If I change it up a bit and do this:

boost::array<char,100> myArr;
map<string,boost::any> myMap;
myArr[51] = a; 
myMap["imageData"] = &myArr;

I have no idea how to get the data back out of the map. How can I create a boost::array<char,100> pointer? Can I? Has anyone else been able to do something similar or has any ideas to do something similar in a more elegant way?

Originally, I wanted to store a char[100] into the map but I could find no way to cast it once in the map (In order to use the data in the map, one must cast it to the proper type). A solution that would allow me to get a char[someSize] out of the map would be ideal.

Answer 1

If you were not using boost::any, for instance doing something like:

map<string, vector> myMap;
myMap["imageData"] = vector(100);

the compiler (GCC 4.6 in my case) would automatically avoid the extra copy. For some reason, boost::any is preventing that. One possible solution would be to use a shared_ptr :

typedef boost::array<char, 100> Image;
map<string, boost::any> myMap;
myMap["imageData"] = boost::shared_ptr<Image>(new Image());

Using a shared_ptr in a container tends to be a much better solution that using a real pointer as you were suggesting in your question. The image would then be destroyed when there were no variable pointing at it, thus saving the programmer from manually freeing its memory.

Depending on the size of your images, however, the overhead of having two copies until the local variable goes out of scope may not be that high. So you should also consider the possibility of keeping your current solution.