Re: invalid use of auto in expression ??

On 5/8/2012 7:12 PM, mdcb808@gmail.com wrote:


I rarely use that form, and I don't remember whether 'obj' is still
declared in the 'else' block or not, so I would actually write (if it
were possible):

    if ((auto obj = mymap.find(key) != mymap.end())
    {
       obj.do_work();
    }
    else
    {
       // deal with it
    }

The more common idiom *I* use involves a pointer:

    if (SomeType* pBlah = getThePointerSomehow())
    {
       // use the non-null pBlah
    }
    else
    {
       // even if 'pBlah' is still declared here, it's null,
       // so no reason to involve that name
    }

which should explain why I don't care to learn whether 'pBlah' would be
still declared in the 'else' part :-)

Now, with your 'find/end' stuff, the only way I see to combine the
declaration/definition/initialization is to declare your own wrapper for
the "find" operation:

    template<class C> struct find_wrapper {
       C& co;
       typename C::iterator it;
       template<class W> find_wrapper(C& c, const W& w)
          : co(c), it(c.find(w)) {}
       // here is the key thing that allows declaring it in an 'if'
       operator bool() const { return it != co.end(); }
       typename C::iterator::value_type& obj() { return *it; }

       void operator =(const find_wrapper&) {}
    };

    template<class C, class W>
    find_wrapper<C> do_find(C& co, const W& w) {
       return find_wrapper<C>(co, w);
    }

    #include <map>

    int main()
    {
       std::map<int,double> mymap;
       int key = 666;

       mymap.find(key);

       if (auto finder = do_find(mymap, key))
       {
          // use finder.obj() somehow:
          std::pair<int,double> id = finder.obj();
       }
       else
       {
          // deal with it
       }
    }

Feel free to modify it.

Victor
--
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