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Home/ Questions/Q 563267
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Editorial Team
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Editorial Team
Asked: May 13, 20262026-05-13T12:38:31+00:00 2026-05-13T12:38:31+00:00

I’m trying to insert some pair value into a map. May map is composed

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I’m trying to insert some pair value into a map. May map is composed by an object and a vector of another object. i don’t know why but the only way to make the code to compile is to declare the first object like a pointer. But in this way when I insert some object, only the first pair is put into the map.

My map is this:

map<prmEdge,vector<prmNode> > archi;

this is the code:

{

bool prmPlanner::insert_edge(int from,int to,int h) {

prmEdge e; 
int f=from; 
int t=to; 
if(to<from){
    f=to;
    t=from; 
} 

e.setFrom(f);
e.setTo(t);

vector<prmNode> app;

prmNode par=nodes[e.getFrom()]; 
prmNode arr=nodes[e.getTo()];

app.push_back(par);
app.push_back(arr);

archi.insert(pair<prmEdge,vector<prmNode> >(e,app) );

return true;
 }

}

In this way, I have an error in compilation in the class pair.h.
What could I do?? Thank you very much.

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  1. Editorial Team
    Editorial Team
    2026-05-13T12:38:31+00:00Added an answer on May 13, 2026 at 12:38 pm

    You need to supply a comparator for prmEdge. My guess is that it uses the default comparator for map, e.g. comparing the address of the key — which is always the same because e is local.

    Objects that serve as Keys in the map need to be ordered, so you either need to supply a operator for comparing edges, or a comparator function for map.

    class EdgeComparator {
    public:
       bool operator( )( const prmEdge& emp1, const prmEdge& emp2) const {
          // ... ?
       }
    };
    
    map<prmEdge,vector<prmNode>, EdgeComparator > archi;
    

    The really hard part is deciding how to compare the edges so a definitive order is defined. Assuming that you only have from and to You can try with:

    class EdgeComparator {
    public:
       bool operator( )( const prmEdge& emp1, const prmEdge& emp2) const {
          if ( emp1.from != emp2.from ) 
              return ( emp1.from < emp2.from );
          return ( emp1.to < emp2.to );
       }
    };
    

    It will sort on primary key from and secondary to.

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