CODEWITHSUNDEEP
c++pointersdictionary
Karl
Karl

Reputation: 5723

Set key and value of a map* in C++

I am following http://www.cplusplus.com/reference/map/map/map/.

I am trying to build an invert index structure, which is a map that has a 64 bits integer as a key and each key will hold a pointer to a sub-map. A sub-map will contain int int pair. So I got myself writing some samples:

map<unsigned long long int, map<int, int>*> invertID;

int main(int argc, char *argv[]){

    map<int,int>* m = new map<int,int>();

    m[10] = 1;

    invertID[1] = m;

    return 0;

}

But here is the trouble, normally, for a non-pointer type map like

std::map<char,int> first;

as described in http://www.cplusplus.com/reference/map/map/map/, I see that we can do 

first['a']= 10;

But if we have a pointer type for map, how can we do that? My code above will generate an error complaining 

error C2679: binary '=' : no operator found which takes a right-hand operand of type 'int' (or there is no acceptable conversion)

Upvotes: 3

Views: 5419

Answers (2)

blackmath
blackmath

Reputation: 252

Map is a red-black tree where operations so you use that to get best performance for search and insert..etc. You are using map not Multi-map in invertID. Thus when you fetch certain key in most cases I would assume you looking for pair of elements. unless you have a need for second map for whatever reason I am not thinking of now, I would go for this:

std::map < unsigned long long int, std::pair < int , int >  > invertID;
int main(){
    std::pair<int,int>  m;
    m = std::make_pair(1,2);
    invertID[1] = m;
    return 0;
}

Have fun.

Upvotes: 2

juanchopanza
juanchopanza

Reputation: 227390

You can de-reference the pointer:

 (*m)[10] = 1;

or

m->operator[](10) = 1;

On the other hand, do you really need all these pointers? Would your program suffer greatly from using this form?

map<unsigned long long int, map<int, int>> invertID;
int main()
{
    map<int,int> m;
    invertID[1][10] = 1;
}

Upvotes: 7

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