object as key to map becomes const in cpp

Question

can someone please explain why the following code compilation fails with message "passing ‘const apple’ as ‘this’ argument of ‘int apple::foo()’ discards qualifiers", and how to resolve it.

#include <cstdlib>
#include <iostream>
#include <string>
#include <map>

using namespace std;

/*
 * 
 */
class apple{
private:
    int a,b,c,d;
public:
    int foo(){
        return a+b+c+d;
    }
};
class ball{
private:
    map<apple,string> mp;
public:
    void foo2(){
        for(map<apple,string>::iterator it = mp.begin();it!=mp.end();++it){
            cout<<it->first.foo()<<endl;
        }
    }

}
int main(int argc, char** argv) {

    return 0;
}

This is not your real code due to a missing ; that terminates your class declaration. Here are the errors from the code you posted — PaulMcKenzie
– PaulMcKenzie, Commented Feb 9, 2018 at 21:51
The map relies on key values to order itself. It can't let you modify them in place, to maintain its own invariants. — StoryTeller - Unslander Monica
– StoryTeller - Unslander Monica, Commented Feb 9, 2018 at 21:52
StoryTeller's right. You should make it int foo() const so you can call it despite the apple being const - which will work for you because foo() doesn't try to change any of the member data in the apple: it should have been const in the first place. — Tony Delroy
– Tony Delroy, Commented Feb 9, 2018 at 22:27

karol · Accepted Answer · 2018-02-09 22:32:24Z

4

Works for me: (added const at the end of foo() and ; on end of ball class). Class apple is a Key in std::map which is declared as const: typedef pair value_type; so accessing key should be also declared as const.

#include <map>
#include <iostream>

using namespace std;

class apple{
 private:
    int a,b,c,d;
public:
    int foo() const {
        return a+b+c+d;
}
};

class ball{
  private:
  map<apple,string> mp;
public:
    void foo2(){
        for(map<apple,string>::iterator it =   mp.begin();it!=mp.end();++it){
         cout<<it->first.foo()<<endl;
    }
}

};

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

    return 0;
}

edited Feb 9, 2018 at 22:32

answered Feb 9, 2018 at 21:58

karol

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