printing values from multimap

[amma]

Hello again,

This little bug has been plaguing me for some time...

I have a self defined type Derivation. The operator << has been overloaded and working elsewhere in the program.

Now I have this piece of code

void ProductionRuleContainer::addProduction(std::string NT, std::string deriv)
{
    Derivation d;
    d.addDerivString(deriv);
    std::cout << "Entering Production Rule:" << d << std::endl;
    prcontainer.insert(std::pair<std::string, Derivation>(NT, d));
    std::cout << NT << ": "<< prcontainer.count(NT) <<  std::endl;
}

This performs exactly the way I want it. It shows the output of d the way I want, before entering in the multimap and the counter works perfectly. The prcontainer multimap is defined as followed:

std::multimap<std::string, Derivation> prcontainer;

Now comes the trouble, I try printing the multimap using the following code:

void ProductionRuleContainer::print() {
    std::cout << ".PRULE" << "\n";
    std::multimap<std::string, Derivation>::iterator it;
    for (it = prcontainer.begin(); it != prcontainer.end();
	  ++it)
	   {
	       std::cout << (*it).first << "\t::==\t" << (*it).second  << std::endl;
	   }
    std::cout << ".END_PRULE" << "\n";
}

It prints the key no problem, which is just a string. But it doesn't print the corresponding derivation value, of which << is loaded and working elsewhere. What am I doing wrong?
Do I need to add a comparer or something for the derivation?
If so how and where must I define this...

Thank you in advance

[Galik]

Can you post the error message?

You could also try modifying it slightly like this to narrow down the error?

void ProductionRuleContainer::print() {
    std::cout << ".PRULE" << "\n";
    std::multimap<std::string, Derivation>::iterator it;
    for (it = prcontainer.begin(); it != prcontainer.end();
	  ++it)
	   {
	       Derivation& d = (*it).second;
	       std::cout << (*it).first << "\t::==\t" << d  << std::endl;
	   }
    std::cout << ".END_PRULE" << "\n";
}

[Galik]

Also can you show us how you defined the operator<< for the Derivation class?

[amma]

hello,

I have adjusted the code and it produces the same results.
there is no error message, it just doesn't print what I want...
The key is printed (a string) and then the tabs and "::==" but It will not print the derivation.

the derivation operator << is really very simple.

std::ostream & operator<<(std::ostream& output, const Derivation& data) {
    return output << data.getOString();
}

This is the header file for derivation

class Derivation {
public:
    Derivation();
    Derivation(const Derivation& orig);
    virtual ~Derivation();
    friend std::ostream & operator<<(std::ostream&, const Derivation&);
    void addElement(std::string x);
    void addDerivString(std::string x);
    std::string getElementId(int entryNo);

private:

    std::string ostring;
    const std::string& getOString() const {
        return ostring;
    }


};

As you see << is no more than printing the ostring.

On line 5, of the first post (addProduction), I am able to print d using << , without problems.
When calling print which invokes << from a derivation iterating through the multimap, it doesn't work.
As I can print the derivation correctly before adding inclines me to believe that the problem occurs when adding to or even when invoking the multimap in the first place.

thank you

[Galik]

Does the code in your copy constructor copy the string 'ostring' over?

[Galik]

You could also test if the Derivation object was not copied correctly by checking it immediately after insertion:

void ProductionRuleContainer::addProduction(std::string NT, std::string deriv)
{
    Derivation d;
    d.addDerivString(deriv);
    std::cout << "Entering Production Rule:" << d << std::endl;
    std::multimap<std::string, Derivation>::iterator it = prcontainer.insert(std::pair<std::string, Derivation>(NT, d));
    std::cout << "After insertion: " << *it << std::endl; // check if it was coppied in okay

    std::cout << NT << ": "<< prcontainer.count(NT) <<  std::endl;
}

[amma]

Yes, problem solved!

Thank you!