c++ - Basic RValue method return -

assuming basic method:

std::string stringtest() {     std::string hello_world("testing...");      return std::move(hello_world); } 

how should use output:

option a:

auto& string_test_output=stringtest(); 

option b:

auto string_test_output=stringtest(); 

is stringtest() temporary value? if so, option not safe. if it's not temporary value, have feeling option b cause copy.

i still getting used rvalue references returning value still scary don't want copying occur or run unnecessary copying!

the best way change method this:

std::string stringtest() {     std::string hello_world("testing...");      return hello_world; } 

there no need tell function returned object should rvalue, comes automatically value being temporary - is, not bound name @ call site. such, initialized value move-constructed (unless temporary inside function elided entirely), no need manually express such intention.

also, since return type std::string, return value, not rvalue-reference, there no point moving return argument before returning. in fact, calling std::move have no effect other possibly tricking compiler perform additional unnecessary move-constructor before returning value. such trick serves no purpose other making code more complex , possibly slower run.

you want return value. 1 reason because of return value optimization (rvo). compilers this, means you're better off returning "copy" gets elided trying clever , returning reference.

even when rvo cannot applied, returning rvalue-reference still not change how returned. when value returned, since temporary @ call site, returned value rvalue can passed along rvalue reference whatever function needs argument. instance, if use return value intialize variable , return value not elided, move constructor still called if available. so, not lost returning value.

given this, should 1 of these catch value:

auto string_test_output = stringtest(); std::string string_test_output = stringtest();