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Does it make sense for a function to return an rvalue reference?

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Does it make sense for a function to return an rvalue reference?


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12















What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?









share|improve this question




















  • 3





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    7 hours ago








  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    7 hours ago






  • 1





    See this answer.

    – lubgr
    7 hours ago











  • Isn't that a forwarding reference?

    – user2357112
    1 hour ago


















12















What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?









share|improve this question




















  • 3





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    7 hours ago








  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    7 hours ago






  • 1





    See this answer.

    – lubgr
    7 hours ago











  • Isn't that a forwarding reference?

    – user2357112
    1 hour ago














12












12








12


3






What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?









share|improve this question
















What would be a valid use case for a signature like this?:



T&& foo();


Or is the rvalue ref only intended for use as argument?



How would one use a function like this?



T&& t = foo(); // is this a thing? And when would t get destructed?






c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited 27 mins ago









Boann

37.7k1291123




37.7k1291123










asked 7 hours ago









Martin B.Martin B.

802313




802313








  • 3





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    7 hours ago








  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    7 hours ago






  • 1





    See this answer.

    – lubgr
    7 hours ago











  • Isn't that a forwarding reference?

    – user2357112
    1 hour ago














  • 3





    std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

    – François Andrieux
    7 hours ago








  • 1





    @FrançoisAndrieux the std::get family of functions, too.

    – Brian
    7 hours ago






  • 1





    See this answer.

    – lubgr
    7 hours ago











  • Isn't that a forwarding reference?

    – user2357112
    1 hour ago








3




3





std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

– François Andrieux
7 hours ago







std::move comes to mind. It certainly returns T&&. Edit : std::optional::value also has an T&& overload. Edit 2 : It also has a const T && overload, though I'll admit I don't understand the meaning.

– François Andrieux
7 hours ago






1




1





@FrançoisAndrieux the std::get family of functions, too.

– Brian
7 hours ago





@FrançoisAndrieux the std::get family of functions, too.

– Brian
7 hours ago




1




1





See this answer.

– lubgr
7 hours ago





See this answer.

– lubgr
7 hours ago













Isn't that a forwarding reference?

– user2357112
1 hour ago





Isn't that a forwarding reference?

– user2357112
1 hour ago












3 Answers
3






active

oldest

votes


















8














For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() { return bar; }
};


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() & { return bar; }
std::vector<int>&& get_vec() && { return std::move(bar); }
};


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer
























  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    7 hours ago











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    7 hours ago











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    7 hours ago






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    6 hours ago








  • 1





    std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    6 hours ago





















2















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.





But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper {

auto operator*() & -> Heavy& {
return heavy;
}

auto operator*() && -> Heavy&& {
return std::move(heavy);
}

private:
Heavy instance;
};

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer


























  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago











  • @CruzJean corrected

    – Guillaume Racicot
    6 hours ago



















0














I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger
{
public:
void log(const char* msg){
logs.append(msg);
}
std::vector<std::string>&& dumpLogs(){
return std::move(logs);
}
private:
std::vector<std::string> logs;
};


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs(){
auto dumped_logs = logs;
return dumped_logs;
}





share|improve this answer
























  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago













  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    6 hours ago











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    4 hours ago












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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









8














For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() { return bar; }
};


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() & { return bar; }
std::vector<int>&& get_vec() && { return std::move(bar); }
};


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer
























  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    7 hours ago











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    7 hours ago











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    7 hours ago






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    6 hours ago








  • 1





    std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    6 hours ago


















8














For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() { return bar; }
};


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() & { return bar; }
std::vector<int>&& get_vec() && { return std::move(bar); }
};


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer
























  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    7 hours ago











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    7 hours ago











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    7 hours ago






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    6 hours ago








  • 1





    std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    6 hours ago
















8












8








8







For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() { return bar; }
};


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() & { return bar; }
std::vector<int>&& get_vec() && { return std::move(bar); }
};


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.






share|improve this answer













For a free function it doesn't make much sense to return a rvalue reference. If it is a non-static local object then you never want to return a reference or pointer to it because it will be destroyed after the function returns. It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not.



One thing that can greatly benefit from returning an rvalue reference is a member function of a temporary object. Lets say you have



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() { return bar; }
};


If you do



auto vec = foo(10).get_vec();


you have to copy because get_vec returns an lvalue. If you instead use



class foo
{
std::vector<int> bar
public:
foo(int n) : bar(n) {}
std::vector<int>& get_vec() & { return bar; }
std::vector<int>&& get_vec() && { return std::move(bar); }
};


Then vec would be able to move the vector returned by get_vec and you save yourself an expensive copy operation.







share|improve this answer












share|improve this answer



share|improve this answer










answered 7 hours ago









NathanOliverNathanOliver

101k16139221




101k16139221













  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    7 hours ago











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    7 hours ago











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    7 hours ago






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    6 hours ago








  • 1





    std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    6 hours ago





















  • @FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

    – NathanOliver
    7 hours ago











  • Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

    – super
    7 hours ago











  • @super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

    – NathanOliver
    7 hours ago






  • 1





    @NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

    – Cruz Jean
    6 hours ago








  • 1





    std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

    – Yakk - Adam Nevraumont
    6 hours ago



















@FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

– NathanOliver
7 hours ago





@FrançoisAndrieux That's covered by my It can possibly make sense to return a rvalue reference to an object that you passed to the function though. It really depends on the use case for if it makes sense or not. catch all. I know there are cases but I really didn't want to try and list them all.

– NathanOliver
7 hours ago













Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

– super
7 hours ago





Is there a reason to prefer returning by rvalue-ref compared to returning by value here?

– super
7 hours ago













@super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

– NathanOliver
7 hours ago





@super I take it your talking about the get_vec case? If you return by value you incur 2 move operations. Passing by rvalue reference you only have 1 move.

– NathanOliver
7 hours ago




1




1





@NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

– Cruz Jean
6 hours ago







@NathanOliver Well, you can call it on any rvalue (e.g. std::move(a).get_vec()). My point is that you're potentially returning a reference to an object that's about to be destroyed. It's the same problem as returning a reference to a local function variable.

– Cruz Jean
6 hours ago






1




1





std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

– Yakk - Adam Nevraumont
6 hours ago







std::vector<int> get_vec() && { return std::move(bar); } in this case ends up being better 999/1000. Can you come up with a better example?

– Yakk - Adam Nevraumont
6 hours ago















2















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.





But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper {

auto operator*() & -> Heavy& {
return heavy;
}

auto operator*() && -> Heavy&& {
return std::move(heavy);
}

private:
Heavy instance;
};

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer


























  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago











  • @CruzJean corrected

    – Guillaume Racicot
    6 hours ago
















2















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.





But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper {

auto operator*() & -> Heavy& {
return heavy;
}

auto operator*() && -> Heavy&& {
return std::move(heavy);
}

private:
Heavy instance;
};

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer


























  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago











  • @CruzJean corrected

    – Guillaume Racicot
    6 hours ago














2












2








2








T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.





But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper {

auto operator*() & -> Heavy& {
return heavy;
}

auto operator*() && -> Heavy&& {
return std::move(heavy);
}

private:
Heavy instance;
};

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());





share|improve this answer
















T&& t = foo(); // is this a thing? And when would t get destructed?



An rvalue reference is really similar to a lvalue reference. Think about your example like it was normal references:



T& foo();

T& t = foo(); // when is t destroyed?


The answer is that t is still valid to use as long as the object is refers to lives.



The same answer still applies to you rvalue reference example.





But... does it make sense to return an rvalue reference?



Sometimes, yes. But very rarely.



consider this:



std::vector<int> v = ...;

// type is std::tuple<std::vector<int>&&>
auto parameters = std::forward_as_tuple(std::move(v));

// fwd is a rvalue reference since std::get returns one.
// fwd is valid as long as v is.
decltype(auto) fwd = std::get<0>(parameters);

// useful for calling function in generic context without copying
consume(std::get<0>(parameters));


So yes there are example. Here, another interesting one:



struct wrapper {

auto operator*() & -> Heavy& {
return heavy;
}

auto operator*() && -> Heavy&& {
return std::move(heavy);
}

private:
Heavy instance;
};

// by value
void use_heavy(Heavy);

// since the wrapper is a temporary, the
// Heavy contained will be a temporary too.
use_heavy(*make_wrapper());






share|improve this answer














share|improve this answer



share|improve this answer








edited 6 hours ago

























answered 7 hours ago









Guillaume RacicotGuillaume Racicot

16.8k53874




16.8k53874













  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago











  • @CruzJean corrected

    – Guillaume Racicot
    6 hours ago



















  • In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago











  • @CruzJean corrected

    – Guillaume Racicot
    6 hours ago

















In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
6 hours ago





In the last example, if the wrapper instance you use this on is a temporary, operator*() returns a reference to instance, which is likewise a temporary. So after the function returns you have a reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
6 hours ago













@CruzJean corrected

– Guillaume Racicot
6 hours ago





@CruzJean corrected

– Guillaume Racicot
6 hours ago











0














I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger
{
public:
void log(const char* msg){
logs.append(msg);
}
std::vector<std::string>&& dumpLogs(){
return std::move(logs);
}
private:
std::vector<std::string> logs;
};


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs(){
auto dumped_logs = logs;
return dumped_logs;
}





share|improve this answer
























  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago













  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    6 hours ago











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    4 hours ago
















0














I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger
{
public:
void log(const char* msg){
logs.append(msg);
}
std::vector<std::string>&& dumpLogs(){
return std::move(logs);
}
private:
std::vector<std::string> logs;
};


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs(){
auto dumped_logs = logs;
return dumped_logs;
}





share|improve this answer
























  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago













  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    6 hours ago











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    4 hours ago














0












0








0







I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger
{
public:
void log(const char* msg){
logs.append(msg);
}
std::vector<std::string>&& dumpLogs(){
return std::move(logs);
}
private:
std::vector<std::string> logs;
};


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs(){
auto dumped_logs = logs;
return dumped_logs;
}





share|improve this answer













I think a use case would be to explicitly give permission to "empty" some non-local variable. Perhaps something like this:



class Logger
{
public:
void log(const char* msg){
logs.append(msg);
}
std::vector<std::string>&& dumpLogs(){
return std::move(logs);
}
private:
std::vector<std::string> logs;
};


But I admit I made this up now, I never actually used it and it also can be done like this:



std::vector<std::string> dumpLogs(){
auto dumped_logs = logs;
return dumped_logs;
}






share|improve this answer












share|improve this answer



share|improve this answer










answered 7 hours ago









QuimbyQuimby

1,272514




1,272514













  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago













  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    6 hours ago











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    4 hours ago



















  • That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

    – Cruz Jean
    6 hours ago













  • @CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

    – Quimby
    6 hours ago











  • return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

    – Cruz Jean
    4 hours ago

















That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
6 hours ago







That's what the other answers have come up with as well, but in the case where this is called on a temporary object, by the time the function returns you have an (rvalue) reference to an object whose lifetime has ended (undefined behavior).

– Cruz Jean
6 hours ago















@CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

– Quimby
6 hours ago





@CruzJean Well, my answer was here first and no one downvoted/complained yet so I will keep it here. I am not returning any reference to a temporary anywhere.

– Quimby
6 hours ago













return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

– Cruz Jean
4 hours ago





return std::move(logs); where the return value is a reference type and *this is a temporary (due to being an rvalue method) makes this->logs a temporary as well. That's the reference to a temporary I mean.

– Cruz Jean
4 hours ago


















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