Does it make sense for a function to return an rvalue reference?Is returning by rvalue reference more...
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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?
Is returning by rvalue reference more efficient?What should main() return in C and C++?Advantages of using forwardClasses, Rvalues and Rvalue ReferencesImage Processing: Algorithm Improvement for 'Coca-Cola Can' RecognitionWhy should I use a pointer rather than the object itself?Does this rvalue signature pattern make sense?Replacing a 32-bit loop counter with 64-bit introduces crazy performance deviationsReturn value or rvalue reference?Rvalue and Lvalue Referencesrvalue reference or forwarding reference?
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}
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++
add a comment |
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++
3
std::move
comes to mind. It certainly returnsT&&
. Edit :std::optional::value
also has anT&&
overload. Edit 2 : It also has aconst T &&
overload, though I'll admit I don't understand the meaning.
– François Andrieux
7 hours ago
1
@FrançoisAndrieux thestd::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
add a comment |
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++
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++
c++
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 returnsT&&
. Edit :std::optional::value
also has anT&&
overload. Edit 2 : It also has aconst T &&
overload, though I'll admit I don't understand the meaning.
– François Andrieux
7 hours ago
1
@FrançoisAndrieux thestd::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
add a comment |
3
std::move
comes to mind. It certainly returnsT&&
. Edit :std::optional::value
also has anT&&
overload. Edit 2 : It also has aconst T &&
overload, though I'll admit I don't understand the meaning.
– François Andrieux
7 hours ago
1
@FrançoisAndrieux thestd::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
add a comment |
3 Answers
3
active
oldest
votes
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.
@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 theget_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
|
show 2 more comments
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());
In the last example, if thewrapper
instance you use this on is a temporary,operator*()
returns a reference toinstance
, 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
add a comment |
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;
}
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) makesthis->logs
a temporary as well. That's the reference to a temporary I mean.
– Cruz Jean
4 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
active
oldest
votes
active
oldest
votes
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.
@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 theget_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
|
show 2 more comments
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.
@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 theget_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
|
show 2 more comments
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.
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.
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 theget_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
|
show 2 more comments
@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 theget_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
|
show 2 more comments
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());
In the last example, if thewrapper
instance you use this on is a temporary,operator*()
returns a reference toinstance
, 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
add a comment |
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());
In the last example, if thewrapper
instance you use this on is a temporary,operator*()
returns a reference toinstance
, 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
add a comment |
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());
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());
edited 6 hours ago
answered 7 hours ago
Guillaume RacicotGuillaume Racicot
16.8k53874
16.8k53874
In the last example, if thewrapper
instance you use this on is a temporary,operator*()
returns a reference toinstance
, 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
add a comment |
In the last example, if thewrapper
instance you use this on is a temporary,operator*()
returns a reference toinstance
, 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
add a comment |
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;
}
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) makesthis->logs
a temporary as well. That's the reference to a temporary I mean.
– Cruz Jean
4 hours ago
add a comment |
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;
}
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) makesthis->logs
a temporary as well. That's the reference to a temporary I mean.
– Cruz Jean
4 hours ago
add a comment |
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;
}
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;
}
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) makesthis->logs
a temporary as well. That's the reference to a temporary I mean.
– Cruz Jean
4 hours ago
add a comment |
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) makesthis->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
add a comment |
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3
std::move
comes to mind. It certainly returnsT&&
. Edit :std::optional::value
also has anT&&
overload. Edit 2 : It also has aconst 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