c++ - std::string and move_iterator -

i writing tokenizer split string , put each of fields inside vector. idea use string::find repeatedly. instead of using temporary string object, used move_iterators, supposed original string see characters stolen algorithm processed it. didn't happen.

this example code demonstrates i'm talking about:

#include <vector> #include <string> #include <iostream>  using namespace std;  void print_strings     ( const vector<string> & v ) {     unsigned int = 1;     ( const auto & s : v )         cout << "#" << i++ << "\t: \"" << s << "\"" << endl;     return; }  int main     ( void ) {     string base( "hello, example string, icescreams" );      /* vector populate strings */     vector<string> v;      /* 1: copy of 'base' */     v.emplace_back( base );     /* 2: copy of 'base' using iterators */     v.emplace_back( base.begin() , base.end() );     /* 3: string think _should_ move 'base' */     v.emplace_back( make_move_iterator(base.begin()) , make_move_iterator(base.end()) );      /* print strings twice      * can see if has changed. */     print_strings( v );     print_strings( v );      return 0; } 

when compiled g++ -std=c++11 -wall -wextra -werror -o2, shows no warnings.

my guessings string's constructors, in range version, copies specified range. i'm not sure, i'd sure and, of course, see workarounds had used.

best regards, kalrish

iterators don't know container.

a move_iterator can't magically move string. can't move underlying element, that's single char , moving char same copying it. need use std::move(base).

#include <vector> #include <string> #include <iostream>  using namespace std;  void print_strings     ( const vector<string> & v ) {     unsigned int = 1;     ( const auto & s : v )         cout << "#" << i++ << "\t: \"" << s << "\"" << endl;     return; }  int main     ( void ) {     string base( "hello, example string, icescreams" );      /* vector populate strings */     vector<string> v;      /* 1: copy of 'base' */     v.emplace_back( base );     /* 2: copy of 'base' using iterators */     v.emplace_back( base.begin() , base.end() );     /* 3: string think _should_ move 'base' */      std::cout << base << '\n'; // base still untouched here      v.emplace_back( std::move(base) ); // it'll moved      print_strings( v );     std::cout << "base: " << base << "/base\n"; // base empty     return 0; } 

see live here.