Efficiency of C++11 push_back() with std::move versus emplace_back() for already constructed objects
Let's see what the different calls that you provided do:
emplace_back(mystring)
: This is an in-place construction of the new element with whatever argument you provided. Since you provided an lvalue, that in-place construction in fact is a copy-construction, i.e. this is the same as callingpush_back(mystring)
push_back(std::move(mystring))
: This calls the move-insertion, which in the case ofstd::string
is an in-place move-construction.emplace_back(std::move(mystring))
: This is again an in-place construction with the arguments you provided. Since that argument is an rvalue, it calls the move-constructor ofstd::string
, i.e. it is an in-place move-construction like in 2.
In other words, if called with one argument of type T, be it an rvalue or lvalue, emplace_back
and push_back
are equivalent.
However, for any other argument(s), emplace_back
wins the race, for example with a char const*
in a vector<string>
:
emplace_back("foo")
callsstd::string(char const*)
for in-place-construction.push_back("foo")
first has to callstd::string(char const*)
for the implicit conversion needed to match the function's signature, and then a move-insertion like case 2. above. Therefore it is equivalent topush_back(string("foo"))