Re: Strange thing while using templates

"Alf P. Steinbach" <>
Sat, 08 Sep 2007 17:51:45 +0200

I have this piece of code:

#include <map>

template <class T> class dict

  std::map<DWORD, T> m_d;

[ ... ]

  void enum_keys_as_dwords (DWORD *out)
    DWORD *o=out;

    for (std::map<DWORD, T>::iterator it=m_d.begin(); it!=m_d.end(); it+

While compiling this in MinGW, it says:

dict:91: error: expected `;' before "it"

(line 91 is the line where "for (...)" is located).

I'm really don't know what to do. The problem is probably in using
type T. Where I mistaken?

std::map<DWORD,T>::iterator could in principle be anything, depending on
the type T. So you need to inform the compiler that it's a type. You
do that by adding the word 'typename' in front.

That said, it would be a good idea to use a std::vector instead of a raw
array, and also to make that pure accessor a 'const' member function
(probably 'DWORD' is a Windows API name, but in general, reserve all
uppercase for macros).

Then the function would look like

   void get_dwords( std::vector<DWORD>& result ) const
       typedef typename std::map<DWORD, T>::const_iterator Iterator;
       std::vector<DWORD> words;
       for( Iterator it = m_d.begin(); it != m_d.end(); ++it )
           words.push_back( it->first );
       words.swap( result );

Note that this way is in general more exception safe as well, not just
more safe against buffer overflows, null-pointer and other problems
associated with raw arrays and pointers.

You can also provide a convenience wrapper relying on Return Value
Optimization (RVO), which most relevant compilers provide:

   std::vector<DWORD> dwords() const
       std::vector<DWORD> result;
       get_dwords( result );
       return result;

The naming convention employed here is that a command-like function's
name 'get_dwords' says what it does, not how (e.g. enumeration) it does
it, and that a function-like function's name 'dwords' says what result
it delivers. That makes the calling code easier to read and comprehend.

Cheers, & hth.,

- Alf

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