What do we need std::as_const() for?
C++11 has given us std::add_const; with C++17, we have a new structure - std::as_const(). The former just tacks a const before the type you provide it with. The second one is a proper (template of a) function, not type trait, which seems to do the same - except for when the type is an rvalue-reference, in which case it cannot be used.
I don't quite understand the motivation for providing std::as_const(). Why do we need it in addition to std::add_const?
c++ const c++17 typetraits rationale
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
|
show 2 more comments
C++11 has given us std::add_const; with C++17, we have a new structure - std::as_const(). The former just tacks a const before the type you provide it with. The second one is a proper (template of a) function, not type trait, which seems to do the same - except for when the type is an rvalue-reference, in which case it cannot be used.
I don't quite understand the motivation for providing std::as_const(). Why do we need it in addition to std::add_const?
c++ const c++17 typetraits rationale
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
I like how you ask questions and then answer to yourself in the very same minute. Surprised you haven't accepted your own answer yet.
– Maxim Egorushkin
Nov 23 '18 at 17:19
5
@MaximEgorushkin see meta.stackoverflow.com/q/250204/1639256
– Oktalist
Nov 23 '18 at 17:55
1
@MaximEgorushkin and other readers: This is intentional, and is actually encouraged here on SO. I was actually planning to just ask this, but as I got an idea on how to find the right standard committee proposal paper. I found it and still needed clarification, so I answered what I did understand; then with commenters' help I was able to complete the answer. The result is a useful question and answer.
– einpoklum
Nov 23 '18 at 22:53
3
@MaximEgorushkin: SO have a feature for exactly this. Check it out: "Ask Question", and there's a checkbox at the bottom of the page: "answer your own question". If you have a better answer than einpoklum's one, feel free to add it.
– geza
Nov 23 '18 at 23:05
1
@MaximEgorushkin: If you're suggesting that the reputation is the reward, then - you're assigning it too much significance. It is just a "gameification" mechanism to promote beneficial activity here on the site. I would lie to you if I said I don't get a slight ego boost from being up-voted - but this is not money, or candy, or anything like that.
– einpoklum
Nov 23 '18 at 23:28
|
show 2 more comments
C++11 has given us std::add_const; with C++17, we have a new structure - std::as_const(). The former just tacks a const before the type you provide it with. The second one is a proper (template of a) function, not type trait, which seems to do the same - except for when the type is an rvalue-reference, in which case it cannot be used.
I don't quite understand the motivation for providing std::as_const(). Why do we need it in addition to std::add_const?
c++ const c++17 typetraits rationale
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
C++11 has given us std::add_const; with C++17, we have a new structure - std::as_const(). The former just tacks a const before the type you provide it with. The second one is a proper (template of a) function, not type trait, which seems to do the same - except for when the type is an rvalue-reference, in which case it cannot be used.
I don't quite understand the motivation for providing std::as_const(). Why do we need it in addition to std::add_const?
c++ const c++17 typetraits rationale
c++ const c++17 typetraits rationale
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
edited Dec 25 '18 at 12:40
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
asked Nov 23 '18 at 16:46
einpoklum
33.4k27121234
33.4k27121234
I like how you ask questions and then answer to yourself in the very same minute. Surprised you haven't accepted your own answer yet.
– Maxim Egorushkin
Nov 23 '18 at 17:19
5
@MaximEgorushkin see meta.stackoverflow.com/q/250204/1639256
– Oktalist
Nov 23 '18 at 17:55
1
@MaximEgorushkin and other readers: This is intentional, and is actually encouraged here on SO. I was actually planning to just ask this, but as I got an idea on how to find the right standard committee proposal paper. I found it and still needed clarification, so I answered what I did understand; then with commenters' help I was able to complete the answer. The result is a useful question and answer.
– einpoklum
Nov 23 '18 at 22:53
3
@MaximEgorushkin: SO have a feature for exactly this. Check it out: "Ask Question", and there's a checkbox at the bottom of the page: "answer your own question". If you have a better answer than einpoklum's one, feel free to add it.
– geza
Nov 23 '18 at 23:05
1
@MaximEgorushkin: If you're suggesting that the reputation is the reward, then - you're assigning it too much significance. It is just a "gameification" mechanism to promote beneficial activity here on the site. I would lie to you if I said I don't get a slight ego boost from being up-voted - but this is not money, or candy, or anything like that.
– einpoklum
Nov 23 '18 at 23:28
|
show 2 more comments
I like how you ask questions and then answer to yourself in the very same minute. Surprised you haven't accepted your own answer yet.
– Maxim Egorushkin
Nov 23 '18 at 17:19
5
@MaximEgorushkin see meta.stackoverflow.com/q/250204/1639256
– Oktalist
Nov 23 '18 at 17:55
1
@MaximEgorushkin and other readers: This is intentional, and is actually encouraged here on SO. I was actually planning to just ask this, but as I got an idea on how to find the right standard committee proposal paper. I found it and still needed clarification, so I answered what I did understand; then with commenters' help I was able to complete the answer. The result is a useful question and answer.
– einpoklum
Nov 23 '18 at 22:53
3
@MaximEgorushkin: SO have a feature for exactly this. Check it out: "Ask Question", and there's a checkbox at the bottom of the page: "answer your own question". If you have a better answer than einpoklum's one, feel free to add it.
– geza
Nov 23 '18 at 23:05
1
@MaximEgorushkin: If you're suggesting that the reputation is the reward, then - you're assigning it too much significance. It is just a "gameification" mechanism to promote beneficial activity here on the site. I would lie to you if I said I don't get a slight ego boost from being up-voted - but this is not money, or candy, or anything like that.
– einpoklum
Nov 23 '18 at 23:28
I like how you ask questions and then answer to yourself in the very same minute. Surprised you haven't accepted your own answer yet.
– Maxim Egorushkin
Nov 23 '18 at 17:19
I like how you ask questions and then answer to yourself in the very same minute. Surprised you haven't accepted your own answer yet.
– Maxim Egorushkin
Nov 23 '18 at 17:19
5
5
@MaximEgorushkin see meta.stackoverflow.com/q/250204/1639256
– Oktalist
Nov 23 '18 at 17:55
@MaximEgorushkin see meta.stackoverflow.com/q/250204/1639256
– Oktalist
Nov 23 '18 at 17:55
1
1
@MaximEgorushkin and other readers: This is intentional, and is actually encouraged here on SO. I was actually planning to just ask this, but as I got an idea on how to find the right standard committee proposal paper. I found it and still needed clarification, so I answered what I did understand; then with commenters' help I was able to complete the answer. The result is a useful question and answer.
– einpoklum
Nov 23 '18 at 22:53
@MaximEgorushkin and other readers: This is intentional, and is actually encouraged here on SO. I was actually planning to just ask this, but as I got an idea on how to find the right standard committee proposal paper. I found it and still needed clarification, so I answered what I did understand; then with commenters' help I was able to complete the answer. The result is a useful question and answer.
– einpoklum
Nov 23 '18 at 22:53
3
3
@MaximEgorushkin: SO have a feature for exactly this. Check it out: "Ask Question", and there's a checkbox at the bottom of the page: "answer your own question". If you have a better answer than einpoklum's one, feel free to add it.
– geza
Nov 23 '18 at 23:05
@MaximEgorushkin: SO have a feature for exactly this. Check it out: "Ask Question", and there's a checkbox at the bottom of the page: "answer your own question". If you have a better answer than einpoklum's one, feel free to add it.
– geza
Nov 23 '18 at 23:05
1
1
@MaximEgorushkin: If you're suggesting that the reputation is the reward, then - you're assigning it too much significance. It is just a "gameification" mechanism to promote beneficial activity here on the site. I would lie to you if I said I don't get a slight ego boost from being up-voted - but this is not money, or candy, or anything like that.
– einpoklum
Nov 23 '18 at 23:28
@MaximEgorushkin: If you're suggesting that the reputation is the reward, then - you're assigning it too much significance. It is just a "gameification" mechanism to promote beneficial activity here on the site. I would lie to you if I said I don't get a slight ego boost from being up-voted - but this is not money, or candy, or anything like that.
– einpoklum
Nov 23 '18 at 23:28
|
show 2 more comments
1 Answer
1
active
oldest
votes
"Need" is a strong word... std::as_const exists because it's useful, not strictly necessary. Since it's a function rather than a trait, we can use it to "add const" to actual values rather than to types.
More specifically: Suppose I have some variable my_value and I want to treat it as a const, but not copy it. Before C++17 I would need to write:
static_cast<const MyType&>(my_value)
and if I don't want to specify the type explicitly, it would be:
static_cast<std::add_const_t<std::remove_reference_t<decltype(my_value)>> &>(my_value)
or if you want to get down and dirty, and use C-style casting:
(const decltype(my_value) &) (my_value)
all of which are annoying and verbose.
Instead of these, with C++17 now write std::as_const(my_value) and that's all there is to it.
Notes:
This function is disabled for rvalue references even though it works just fine for them. The reason is to help you avoid inadvertantly keeping a reference to a temporary past its destruction. As @NicolBolas explains, if you write something like:
for(auto &x : std::as_const(returns_container())) { /* do stuff with x */ }
then the returned container's lifetime ends before the first iteration of the loop. Very easy to miss!
For additional (?) information, consult the official proposition of this utility function: P007R1, by Adam David Alan Martin and Alisdair Meredith.
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
2
Worse still: you needstd::remove_reference_tifmy_valueis itself a reference.
– Quentin
Nov 23 '18 at 16:49
1
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to useas_conston them. If you ever callas_conston a temporary, you will get a reference to a destroyed object.
– Nicol Bolas
Nov 23 '18 at 16:53
1
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which theas_const()is used, is fully executed? If not, can you link to an explanation?
– einpoklum
Nov 23 '18 at 16:57
2
@einpoklum: It will life to the end of the full expression. But if you're doing something likefor(auto &val : std::as_const(returns_container())), the "full expression" ends before theforloop starts. Whereas if you had usedreturns_container()directly, the temporary's lifetime would extend to the entireforloop. Better to prevent people from making that mistake up-front.
– Nicol Bolas
Nov 23 '18 at 17:02
1
std::as_const()is particularly useful to select theconstoverload of a set. Givenstd::map<int,int> m,m[0]andas_const(m)[0]are quite different.
– YSC
Nov 23 '18 at 17:14
|
show 9 more comments
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"Need" is a strong word... std::as_const exists because it's useful, not strictly necessary. Since it's a function rather than a trait, we can use it to "add const" to actual values rather than to types.
More specifically: Suppose I have some variable my_value and I want to treat it as a const, but not copy it. Before C++17 I would need to write:
static_cast<const MyType&>(my_value)
and if I don't want to specify the type explicitly, it would be:
static_cast<std::add_const_t<std::remove_reference_t<decltype(my_value)>> &>(my_value)
or if you want to get down and dirty, and use C-style casting:
(const decltype(my_value) &) (my_value)
all of which are annoying and verbose.
Instead of these, with C++17 now write std::as_const(my_value) and that's all there is to it.
Notes:
This function is disabled for rvalue references even though it works just fine for them. The reason is to help you avoid inadvertantly keeping a reference to a temporary past its destruction. As @NicolBolas explains, if you write something like:
for(auto &x : std::as_const(returns_container())) { /* do stuff with x */ }
then the returned container's lifetime ends before the first iteration of the loop. Very easy to miss!
For additional (?) information, consult the official proposition of this utility function: P007R1, by Adam David Alan Martin and Alisdair Meredith.
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
2
Worse still: you needstd::remove_reference_tifmy_valueis itself a reference.
– Quentin
Nov 23 '18 at 16:49
1
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to useas_conston them. If you ever callas_conston a temporary, you will get a reference to a destroyed object.
– Nicol Bolas
Nov 23 '18 at 16:53
1
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which theas_const()is used, is fully executed? If not, can you link to an explanation?
– einpoklum
Nov 23 '18 at 16:57
2
@einpoklum: It will life to the end of the full expression. But if you're doing something likefor(auto &val : std::as_const(returns_container())), the "full expression" ends before theforloop starts. Whereas if you had usedreturns_container()directly, the temporary's lifetime would extend to the entireforloop. Better to prevent people from making that mistake up-front.
– Nicol Bolas
Nov 23 '18 at 17:02
1
std::as_const()is particularly useful to select theconstoverload of a set. Givenstd::map<int,int> m,m[0]andas_const(m)[0]are quite different.
– YSC
Nov 23 '18 at 17:14
|
show 9 more comments
"Need" is a strong word... std::as_const exists because it's useful, not strictly necessary. Since it's a function rather than a trait, we can use it to "add const" to actual values rather than to types.
More specifically: Suppose I have some variable my_value and I want to treat it as a const, but not copy it. Before C++17 I would need to write:
static_cast<const MyType&>(my_value)
and if I don't want to specify the type explicitly, it would be:
static_cast<std::add_const_t<std::remove_reference_t<decltype(my_value)>> &>(my_value)
or if you want to get down and dirty, and use C-style casting:
(const decltype(my_value) &) (my_value)
all of which are annoying and verbose.
Instead of these, with C++17 now write std::as_const(my_value) and that's all there is to it.
Notes:
This function is disabled for rvalue references even though it works just fine for them. The reason is to help you avoid inadvertantly keeping a reference to a temporary past its destruction. As @NicolBolas explains, if you write something like:
for(auto &x : std::as_const(returns_container())) { /* do stuff with x */ }
then the returned container's lifetime ends before the first iteration of the loop. Very easy to miss!
For additional (?) information, consult the official proposition of this utility function: P007R1, by Adam David Alan Martin and Alisdair Meredith.
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
2
Worse still: you needstd::remove_reference_tifmy_valueis itself a reference.
– Quentin
Nov 23 '18 at 16:49
1
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to useas_conston them. If you ever callas_conston a temporary, you will get a reference to a destroyed object.
– Nicol Bolas
Nov 23 '18 at 16:53
1
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which theas_const()is used, is fully executed? If not, can you link to an explanation?
– einpoklum
Nov 23 '18 at 16:57
2
@einpoklum: It will life to the end of the full expression. But if you're doing something likefor(auto &val : std::as_const(returns_container())), the "full expression" ends before theforloop starts. Whereas if you had usedreturns_container()directly, the temporary's lifetime would extend to the entireforloop. Better to prevent people from making that mistake up-front.
– Nicol Bolas
Nov 23 '18 at 17:02
1
std::as_const()is particularly useful to select theconstoverload of a set. Givenstd::map<int,int> m,m[0]andas_const(m)[0]are quite different.
– YSC
Nov 23 '18 at 17:14
|
show 9 more comments
"Need" is a strong word... std::as_const exists because it's useful, not strictly necessary. Since it's a function rather than a trait, we can use it to "add const" to actual values rather than to types.
More specifically: Suppose I have some variable my_value and I want to treat it as a const, but not copy it. Before C++17 I would need to write:
static_cast<const MyType&>(my_value)
and if I don't want to specify the type explicitly, it would be:
static_cast<std::add_const_t<std::remove_reference_t<decltype(my_value)>> &>(my_value)
or if you want to get down and dirty, and use C-style casting:
(const decltype(my_value) &) (my_value)
all of which are annoying and verbose.
Instead of these, with C++17 now write std::as_const(my_value) and that's all there is to it.
Notes:
This function is disabled for rvalue references even though it works just fine for them. The reason is to help you avoid inadvertantly keeping a reference to a temporary past its destruction. As @NicolBolas explains, if you write something like:
for(auto &x : std::as_const(returns_container())) { /* do stuff with x */ }
then the returned container's lifetime ends before the first iteration of the loop. Very easy to miss!
For additional (?) information, consult the official proposition of this utility function: P007R1, by Adam David Alan Martin and Alisdair Meredith.
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
"Need" is a strong word... std::as_const exists because it's useful, not strictly necessary. Since it's a function rather than a trait, we can use it to "add const" to actual values rather than to types.
More specifically: Suppose I have some variable my_value and I want to treat it as a const, but not copy it. Before C++17 I would need to write:
static_cast<const MyType&>(my_value)
and if I don't want to specify the type explicitly, it would be:
static_cast<std::add_const_t<std::remove_reference_t<decltype(my_value)>> &>(my_value)
or if you want to get down and dirty, and use C-style casting:
(const decltype(my_value) &) (my_value)
all of which are annoying and verbose.
Instead of these, with C++17 now write std::as_const(my_value) and that's all there is to it.
Notes:
This function is disabled for rvalue references even though it works just fine for them. The reason is to help you avoid inadvertantly keeping a reference to a temporary past its destruction. As @NicolBolas explains, if you write something like:
for(auto &x : std::as_const(returns_container())) { /* do stuff with x */ }
then the returned container's lifetime ends before the first iteration of the loop. Very easy to miss!
For additional (?) information, consult the official proposition of this utility function: P007R1, by Adam David Alan Martin and Alisdair Meredith.
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
edited Nov 26 '18 at 13:37
Arne Vogel
3,85011125
3,85011125
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
answered Nov 23 '18 at 16:46
einpoklum
33.4k27121234
33.4k27121234
2
Worse still: you needstd::remove_reference_tifmy_valueis itself a reference.
– Quentin
Nov 23 '18 at 16:49
1
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to useas_conston them. If you ever callas_conston a temporary, you will get a reference to a destroyed object.
– Nicol Bolas
Nov 23 '18 at 16:53
1
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which theas_const()is used, is fully executed? If not, can you link to an explanation?
– einpoklum
Nov 23 '18 at 16:57
2
@einpoklum: It will life to the end of the full expression. But if you're doing something likefor(auto &val : std::as_const(returns_container())), the "full expression" ends before theforloop starts. Whereas if you had usedreturns_container()directly, the temporary's lifetime would extend to the entireforloop. Better to prevent people from making that mistake up-front.
– Nicol Bolas
Nov 23 '18 at 17:02
1
std::as_const()is particularly useful to select theconstoverload of a set. Givenstd::map<int,int> m,m[0]andas_const(m)[0]are quite different.
– YSC
Nov 23 '18 at 17:14
|
show 9 more comments
2
Worse still: you needstd::remove_reference_tifmy_valueis itself a reference.
– Quentin
Nov 23 '18 at 16:49
1
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to useas_conston them. If you ever callas_conston a temporary, you will get a reference to a destroyed object.
– Nicol Bolas
Nov 23 '18 at 16:53
1
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which theas_const()is used, is fully executed? If not, can you link to an explanation?
– einpoklum
Nov 23 '18 at 16:57
2
@einpoklum: It will life to the end of the full expression. But if you're doing something likefor(auto &val : std::as_const(returns_container())), the "full expression" ends before theforloop starts. Whereas if you had usedreturns_container()directly, the temporary's lifetime would extend to the entireforloop. Better to prevent people from making that mistake up-front.
– Nicol Bolas
Nov 23 '18 at 17:02
1
std::as_const()is particularly useful to select theconstoverload of a set. Givenstd::map<int,int> m,m[0]andas_const(m)[0]are quite different.
– YSC
Nov 23 '18 at 17:14
2
2
Worse still: you need
std::remove_reference_t if my_value is itself a reference.– Quentin
Nov 23 '18 at 16:49
Worse still: you need
std::remove_reference_t if my_value is itself a reference.– Quentin
Nov 23 '18 at 16:49
1
1
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to use
as_const on them. If you ever call as_const on a temporary, you will get a reference to a destroyed object.– Nicol Bolas
Nov 23 '18 at 16:53
"I'm not entirely clear why the deletion for rvalue references is necessary though." Because C++'s temporary lifetime extension rules don't work correctly if you tried to use
as_const on them. If you ever call as_const on a temporary, you will get a reference to a destroyed object.– Nicol Bolas
Nov 23 '18 at 16:53
1
1
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which the
as_const() is used, is fully executed? If not, can you link to an explanation?– einpoklum
Nov 23 '18 at 16:57
@NicolBolas: Thanks. But - won't the temporary live until the the statement in which the
as_const() is used, is fully executed? If not, can you link to an explanation?– einpoklum
Nov 23 '18 at 16:57
2
2
@einpoklum: It will life to the end of the full expression. But if you're doing something like
for(auto &val : std::as_const(returns_container())), the "full expression" ends before the for loop starts. Whereas if you had used returns_container() directly, the temporary's lifetime would extend to the entire for loop. Better to prevent people from making that mistake up-front.– Nicol Bolas
Nov 23 '18 at 17:02
@einpoklum: It will life to the end of the full expression. But if you're doing something like
for(auto &val : std::as_const(returns_container())), the "full expression" ends before the for loop starts. Whereas if you had used returns_container() directly, the temporary's lifetime would extend to the entire for loop. Better to prevent people from making that mistake up-front.– Nicol Bolas
Nov 23 '18 at 17:02
1
1
std::as_const() is particularly useful to select the const overload of a set. Given std::map<int,int> m, m[0] and as_const(m)[0] are quite different.– YSC
Nov 23 '18 at 17:14
std::as_const() is particularly useful to select the const overload of a set. Given std::map<int,int> m, m[0] and as_const(m)[0] are quite different.– YSC
Nov 23 '18 at 17:14
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I like how you ask questions and then answer to yourself in the very same minute. Surprised you haven't accepted your own answer yet.
– Maxim Egorushkin
Nov 23 '18 at 17:19
5
@MaximEgorushkin see meta.stackoverflow.com/q/250204/1639256
– Oktalist
Nov 23 '18 at 17:55
1
@MaximEgorushkin and other readers: This is intentional, and is actually encouraged here on SO. I was actually planning to just ask this, but as I got an idea on how to find the right standard committee proposal paper. I found it and still needed clarification, so I answered what I did understand; then with commenters' help I was able to complete the answer. The result is a useful question and answer.
– einpoklum
Nov 23 '18 at 22:53
3
@MaximEgorushkin: SO have a feature for exactly this. Check it out: "Ask Question", and there's a checkbox at the bottom of the page: "answer your own question". If you have a better answer than einpoklum's one, feel free to add it.
– geza
Nov 23 '18 at 23:05
1
@MaximEgorushkin: If you're suggesting that the reputation is the reward, then - you're assigning it too much significance. It is just a "gameification" mechanism to promote beneficial activity here on the site. I would lie to you if I said I don't get a slight ego boost from being up-voted - but this is not money, or candy, or anything like that.
– einpoklum
Nov 23 '18 at 23:28