Re: vector<string> to char*[]

From:
"Alf P. Steinbach" <alfps@start.no>
Newsgroups:
comp.lang.c++
Date:
Tue, 02 Oct 2007 11:45:42 +0200
Message-ID:
<13g44q9b9rke9d4@corp.supernews.com>
* Andrew Wingorodov:

Joel Yliluoma <bisqwit@iki.fi> wrote:

 const char** argv = new const char*[ arg.size() ];
 for(size_t a=0; a<arg.size(); ++a) argv[a] = arg[a].c_str();
 execv(... argv ...);
 // cleanup when error happened
 delete[] argv;


i do correct?

//
const char**
 args::constchar (const std::vector<std::string>& arg_)
{
    static std::auto_ptr<const char*> v;
    v = (std::auto_ptr<const char*>) new const char* [arg_.size()+1];

auto_ptr calls delete, whereas you need delete[]. Not that it matters
much in practice for a char array. But formally UB.

     std::vector<std::string>::const_iterator i = arg_.begin();
    size_t j;

    for (; i != arg_.end (); ++i)
    {
        v.get()[j++] = (*i).c_str() ;
    }

    v.get()[j] = NULL;
    return v.get();
}


When the function returns, the auto_ptr calls delete. As mentioned
that's just formally UB, but if that delete works, the effect is that
the function returns a pointer to deallocated memory.

Try what I posted earlier, it's safe:

   std::vector<string> arg;
   ...

   std::vector<char const*> v( arg.size() );
   for( size_t i = 0; i < v.size(); ++i ) { v[i] = arg[i].c_str(); }
   blahblah( &v[0] );

If you have any questions, just ask.

Cheers, & hth.,

- Alf

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