Confusing Template error

Originally Posted by Evil g++

In file included from main.cpp:3:0:
scope.h: In function ‘double mm::get_num(const string&, const mm::scope<std::basic_string<char> >&)’:
scope.h:40:30: error: passing ‘const mm::scope<std::basic_string<char> >’ as ‘this’ argument of ‘T* mm::scope<T>::find(const string&) [with T = std::basic_string<char>, std::string = std::basic_string<char>]’ discards qualifiers [-fpermissive]
make: *** [main.o] Error 1

main.cpp has nothing important now...
Here is the scope.h

Code:

#ifndef SCOPE
#define SCOPE
#include<vector>
#include<map>
#include<sstream>
namespace mm
{
    template<typename T>
    class scope
    {
        std::vector< std::map<std::string,T> > data;
    public:
        T* find(const std::string& s)
        {
            for(auto m=data.rbegin();m!=data.rend();m++)
            {
                auto p = m->find(s);
                if(p!=m->end())return &(p->second);
            }
            return nullptr;
        };
        void new_local(const std::map<std::string,T>& sm){data.push_back(sm);};
        void exit_scope(){data.pop_back();};
    };

    /*Converts a string to a double
     * If the string is an identifier....
     * it fetches the value from the scope object supplied. 
     */
    double get_num(const std::string& s, const scope<std::string>& ss)
    {
        std::string numstr;
        std::istringstream is(s);
        double d;
        if(is>>d)
            return d;
        else 
        {
            std::string* x = ss.find(s);
            if(x!=nullptr)numstr=*x;
            else throw("NAN");//change later
        }
        is.str(numstr);
        if(is>>d)
            return d;
        else throw("NAN");//change later
        
    }
}

#endif

I think the most important part of your error message is "discards qualifiers", this seems to be a const correctness issue.

Jim

Correct you are; have my thanks, you do.

Problem solved by making the scope passed into the get_num not a const.
(Any idea why the error was there when I never modified the object ?)

Jim is correct - specifically, ss is declared const, but find() is not a const method, so you can't call it on ss.

Bah - too late!

Originally Posted by manasij7479

(Any idea why the error was there when I never modified the object ?)

Define "never modified the object". If you mean that find() doesn't modify the object, that's not enough - in order to be able to use it on const-qualified instances of your class, you have to declare find() to be const to let the compiler (a) know that and (b) enforce it within the method.

Originally Posted by JohnGraham

Define "never modified the object". If you mean that find() doesn't modify the object, that's not enough - in order to be able to use it on const-qualified instances of your class, you have to declare find() to be const to let the compiler (a) know that and (b) enforce it within the method.

Yes.. I didn't put const because I'd use the pointer returned by it to change the object (in other cases..not this)
But I now realize that the 'viral-ity' of const won't allow that.

If a function takes a reference or pointer to something and returns a non-const reference or pointer to that same object, then the function itself can be const, but the parameter cannot.
Typically, the solution is to overload both variants.