在 boost 中，如何将 boost 迭代器传递给 function 以某种方式转换为 std::string

Question

See specific question as a comment at the end of the following code.

std::string s("my sample string \"with quotes\"");

boost::escaped_list_separator<char> 
els(""," ","\"\'");

boost::tokenizer<boost::escaped_list_separator<char> >::iterator 
itr;

boost::tokenizer<boost::escaped_list_separator<char> > 
tok(s, els);

itr=tok.begin();
if (itr!=tok.end()) 
    fn_that_receives_pointer_to_std_string(itr); // <---- IS IT POSSIBLE TO SEND POINTER AND NOT HAVE TO CREATE A NEW STRING ??

Answer 1

boost::tokenizer<boost::escaped_list_separator<char> >::iterator is not a pointer to std::string , but you can turn it into std::string const * with

&(*itr)

If a const pointer is not what you must pass, you may be able to do

std::string s(*itr);

and pass &s , depending on the ownership semantics of fn_that_receives_pointer_to_std_string . Boost Tokenizer does no distinguish between iterator and const_iterator , so the result of operator* is always const .

Answer 2

*itr will actually return a basic_string instead of a string , so you need to convert one to another:

using namespace std;
using namespace boost;

void fn_that_receives_pointer_to_std_string(string* str)
{
    cout << "str: " << *str << endl;
}

int main()
{
   string s = "Field 1,\"putting quotes around fields, allows commas\",Field 3";
   tokenizer<escaped_list_separator<char> > tok(s);
   for(tokenizer<escaped_list_separator<char> >::iterator beg=tok.begin(); beg!=tok.end();++beg)
   {   
       string tmp(*beg);
       fn_that_receives_pointer_to_std_string(&tmp);
   }   
}

I don't like the idea to passing the memory address of a string to another function. Consider passing it by copy or by reference.

Answer 3

Sorry, it's impossible.

It's exactly the reason why the rule "take string parameters as std::string " is wrong. boost::iterator_range<const char*> can be better when a template is inappropriate (separate compilation for example).