c++ - How do you convert a lvalue to an rvalue? And what happens to the `new` lvalue? -

i move object std::vector using std::vector::push_back(). seem possible since there std::vector::push_back(value_type&& val) function. due the existence of std::vector::push_back(value_type const & val), copies , overriding call, need convert lvalue object rvalue.

how done?

example:

struct x { int x; x(int x) : x(x) {} };  int main() {     std::vector<x> ax;     x x(3);     ax.push_back(x);  // <= want move x in vector ax, not copy.     return 0; } 

actually, maybe can't be? ask because after writing example, new questions are:

• if move x ax, value of x.x? if x had explicit destructor, happen x when leaves scope?

just use std::move():

ax.push_back(std::move(x)); 

concerning question:

if move x ax, value of x.x?

in case, class x not contain explicitly declared move constructor, compiler generate 1 memberwise move of x's members. since x has 1 member of type int, , moving int no different copying it, x.x have same value had before being moved from.

if x had explicit destructor, happen x when leaves scope?

if x has user-declared destructor, inhibit generation of implicit move constructor - not of implicit copy constructor. therefore, function call:

ax.push_back(std::move(x)); 

will result in x being copied ax. in case, x destroyed when going out of scope, since has automatic storage duration - no matter whether user-declared destructor present or not.

in general, 1 should not make assumptions on state of object has been moved from, except that state valid (paragraph 17.6.5.15/1 of c++11 standard guarantees case types of standard library).

in concrete, means the functions can safely work object has been moved not have precondition on state of object. typically, 2 such functions destructor , (copy- or move-) assignment operator.