CODEWITHSUNDEEP
c++syntaxtemplates
Andrew
Andrew

Reputation: 12009

c++ template syntax error

My C++ is a little rusty having worked in Java and C# for the last half dozen years. I've got a stupid little error that I just cannot figure out.

I've pared the code down as much as possible.

#include <list>
template<class T> class Subscriber
{
    virtual void published( T t ) = 0;
};

template <class T> class PubSub
{
private:
    std::list< Subscriber<T>* > subscribers;
public:
    void publish( T t );
};

template<class T> void PubSub<T>::publish( T t ) 
{
    for( std::list< Subscriber<T>* >::iterator i = subscribers.begin(); i != subscribers.end(); ++i )
        i->published( t );
}

When I try and compile this (by including this header file in a code file), I get the following error:

../util/pubsub.h: In member function ‘void PubSub<T>::publish(T)’:
../util/pubsub.h:18: error: expected `;' before ‘i’
../util/pubsub.h:18: error: ‘i’ was not declared in this scope

What am I missing here?

Upvotes: 2

Views: 938

Answers (3)

Marsh Ray
Marsh Ray

Reputation: 2875

for( typename std::list< Subscriber<T>* >::iterator i = ...
     ^^^^^^^^

Upvotes: 6

Stack Overflow is garbage
Stack Overflow is garbage

Reputation: 248139

for( typename std::list< Subscriber<T>* >::iterator i = subscribers.begin(); i != subscribers.end(); ++i )

You need the typename because iterator is a dependent name. The compiler has to check the template type T before it knows whether iterator is a type or a value. In those cases, it assumes it to be a value, unless you add typename.

Upvotes: 6

Tyler McHenry
Tyler McHenry

Reputation: 76730

This

std::list< Subscriber<T>* >::iterator

needs to be this

typename std::list< Subscriber<T>* >::iterator

The compiler assumes nested names in templates are static variables (not types) until told otherwise.

Upvotes: 4

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