What do we need std::as_const() for?












16















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







share|improve this question

























  • 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
















16















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







share|improve this question

























  • 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














16












16








16


1






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







share|improve this question
















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






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited Dec 25 '18 at 12:40







einpoklum

















asked Nov 23 '18 at 16:46









einpoklumeinpoklum

34.7k27128247




34.7k27128247













  • 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








  • 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












1 Answer
1






active

oldest

votes


















19














"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.







share|improve this answer





















  • 2





    Worse still: you need std::remove_reference_t if my_value is 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 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





    @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





    @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





    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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1 Answer
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1 Answer
1






active

oldest

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active

oldest

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active

oldest

votes









19














"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.







share|improve this answer





















  • 2





    Worse still: you need std::remove_reference_t if my_value is 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 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





    @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





    @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





    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
















19














"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.







share|improve this answer





















  • 2





    Worse still: you need std::remove_reference_t if my_value is 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 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





    @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





    @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





    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














19












19








19







"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.







share|improve this answer















"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.








share|improve this answer














share|improve this answer



share|improve this answer








edited Nov 26 '18 at 13:37









Arne Vogel

4,49521326




4,49521326










answered Nov 23 '18 at 16:46









einpoklumeinpoklum

34.7k27128247




34.7k27128247








  • 2





    Worse still: you need std::remove_reference_t if my_value is 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 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





    @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





    @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





    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














  • 2





    Worse still: you need std::remove_reference_t if my_value is 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 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





    @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





    @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





    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








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




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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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