Can a concept evaluation depend on where it is evaluated?











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[temp.concept]/5 says:




A concept is not instantiated ([temp.spec]).
[ Note: An id-expression that denotes a concept specialization is evaluated as an expression ([expr.prim.id]). [...]]




Does it mean that this rule bellow ([temp.point]/8) does not apply?




If two different points of instantiation give a template specialization different meanings according to the one-definition rule, the program is ill-formed, no diagnostic required.






For example if this rule does not apply, this code bellow is well formed:



template<class T>
concept Complete = sizeof(T)==sizeof(T);

struct A;

constexpr inline bool b1 = Complete<A>; //Complete<A>==false;

struct A{};

constexpr inline bool b2 = Complete<A>; //Complete<A>==true;


This question is followed by this one










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  • @bolov Or template<class A> :-P
    – Zereges
    Nov 12 at 10:59








  • 1




    Are you sure the concept definition is well-formed when applied for b1?
    – rubenvb
    Nov 12 at 11:03










  • That concept is ill-formed upon definition IIUC.
    – StoryTeller
    Nov 12 at 11:05












  • @Oliv - The paragraph I linked says nothing about instantiation, it's about template definition and name lookup. I'd be shocked if one can conjure non-dependent identifiers willy-nilly in a concept.
    – StoryTeller
    Nov 12 at 12:59












  • @Oliv - Yes, sizeof(A). It was an entirely different question with that typo. But whatever, if it was "obvious".
    – StoryTeller
    Nov 12 at 13:01

















up vote
14
down vote

favorite
2












[temp.concept]/5 says:




A concept is not instantiated ([temp.spec]).
[ Note: An id-expression that denotes a concept specialization is evaluated as an expression ([expr.prim.id]). [...]]




Does it mean that this rule bellow ([temp.point]/8) does not apply?




If two different points of instantiation give a template specialization different meanings according to the one-definition rule, the program is ill-formed, no diagnostic required.






For example if this rule does not apply, this code bellow is well formed:



template<class T>
concept Complete = sizeof(T)==sizeof(T);

struct A;

constexpr inline bool b1 = Complete<A>; //Complete<A>==false;

struct A{};

constexpr inline bool b2 = Complete<A>; //Complete<A>==true;


This question is followed by this one










share|improve this question
























  • @bolov Or template<class A> :-P
    – Zereges
    Nov 12 at 10:59








  • 1




    Are you sure the concept definition is well-formed when applied for b1?
    – rubenvb
    Nov 12 at 11:03










  • That concept is ill-formed upon definition IIUC.
    – StoryTeller
    Nov 12 at 11:05












  • @Oliv - The paragraph I linked says nothing about instantiation, it's about template definition and name lookup. I'd be shocked if one can conjure non-dependent identifiers willy-nilly in a concept.
    – StoryTeller
    Nov 12 at 12:59












  • @Oliv - Yes, sizeof(A). It was an entirely different question with that typo. But whatever, if it was "obvious".
    – StoryTeller
    Nov 12 at 13:01















up vote
14
down vote

favorite
2









up vote
14
down vote

favorite
2






2





[temp.concept]/5 says:




A concept is not instantiated ([temp.spec]).
[ Note: An id-expression that denotes a concept specialization is evaluated as an expression ([expr.prim.id]). [...]]




Does it mean that this rule bellow ([temp.point]/8) does not apply?




If two different points of instantiation give a template specialization different meanings according to the one-definition rule, the program is ill-formed, no diagnostic required.






For example if this rule does not apply, this code bellow is well formed:



template<class T>
concept Complete = sizeof(T)==sizeof(T);

struct A;

constexpr inline bool b1 = Complete<A>; //Complete<A>==false;

struct A{};

constexpr inline bool b2 = Complete<A>; //Complete<A>==true;


This question is followed by this one










share|improve this question















[temp.concept]/5 says:




A concept is not instantiated ([temp.spec]).
[ Note: An id-expression that denotes a concept specialization is evaluated as an expression ([expr.prim.id]). [...]]




Does it mean that this rule bellow ([temp.point]/8) does not apply?




If two different points of instantiation give a template specialization different meanings according to the one-definition rule, the program is ill-formed, no diagnostic required.






For example if this rule does not apply, this code bellow is well formed:



template<class T>
concept Complete = sizeof(T)==sizeof(T);

struct A;

constexpr inline bool b1 = Complete<A>; //Complete<A>==false;

struct A{};

constexpr inline bool b2 = Complete<A>; //Complete<A>==true;


This question is followed by this one







c++ language-lawyer c++-concepts c++20






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edited Nov 12 at 13:35

























asked Nov 12 at 10:21









Oliv

8,1351954




8,1351954












  • @bolov Or template<class A> :-P
    – Zereges
    Nov 12 at 10:59








  • 1




    Are you sure the concept definition is well-formed when applied for b1?
    – rubenvb
    Nov 12 at 11:03










  • That concept is ill-formed upon definition IIUC.
    – StoryTeller
    Nov 12 at 11:05












  • @Oliv - The paragraph I linked says nothing about instantiation, it's about template definition and name lookup. I'd be shocked if one can conjure non-dependent identifiers willy-nilly in a concept.
    – StoryTeller
    Nov 12 at 12:59












  • @Oliv - Yes, sizeof(A). It was an entirely different question with that typo. But whatever, if it was "obvious".
    – StoryTeller
    Nov 12 at 13:01




















  • @bolov Or template<class A> :-P
    – Zereges
    Nov 12 at 10:59








  • 1




    Are you sure the concept definition is well-formed when applied for b1?
    – rubenvb
    Nov 12 at 11:03










  • That concept is ill-formed upon definition IIUC.
    – StoryTeller
    Nov 12 at 11:05












  • @Oliv - The paragraph I linked says nothing about instantiation, it's about template definition and name lookup. I'd be shocked if one can conjure non-dependent identifiers willy-nilly in a concept.
    – StoryTeller
    Nov 12 at 12:59












  • @Oliv - Yes, sizeof(A). It was an entirely different question with that typo. But whatever, if it was "obvious".
    – StoryTeller
    Nov 12 at 13:01


















@bolov Or template<class A> :-P
– Zereges
Nov 12 at 10:59






@bolov Or template<class A> :-P
– Zereges
Nov 12 at 10:59






1




1




Are you sure the concept definition is well-formed when applied for b1?
– rubenvb
Nov 12 at 11:03




Are you sure the concept definition is well-formed when applied for b1?
– rubenvb
Nov 12 at 11:03












That concept is ill-formed upon definition IIUC.
– StoryTeller
Nov 12 at 11:05






That concept is ill-formed upon definition IIUC.
– StoryTeller
Nov 12 at 11:05














@Oliv - The paragraph I linked says nothing about instantiation, it's about template definition and name lookup. I'd be shocked if one can conjure non-dependent identifiers willy-nilly in a concept.
– StoryTeller
Nov 12 at 12:59






@Oliv - The paragraph I linked says nothing about instantiation, it's about template definition and name lookup. I'd be shocked if one can conjure non-dependent identifiers willy-nilly in a concept.
– StoryTeller
Nov 12 at 12:59














@Oliv - Yes, sizeof(A). It was an entirely different question with that typo. But whatever, if it was "obvious".
– StoryTeller
Nov 12 at 13:01






@Oliv - Yes, sizeof(A). It was an entirely different question with that typo. But whatever, if it was "obvious".
– StoryTeller
Nov 12 at 13:01














1 Answer
1






active

oldest

votes

















up vote
7
down vote



accepted











Can a concept evaluation depend on where it is evaluated?




Yes. This was explicitly discussed during core wording review when merging Concepts into the working draft. The concept is re-evaluated every time.



As a result, this:



template<class T>
concept Complete = sizeof(T) == sizeof(T);

struct A;
static_assert(!Complete<A>);
struct A {};
static_assert(Complete<A>);


is well-formed. In other words, we don't "memoize" concepts in the same way we "memoize" template instantiations.






share|improve this answer





















  • Thanks! Could you have a look at this question too?
    – Oliv
    Nov 12 at 18:50








  • 1




    Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
    – Davis Herring
    Nov 12 at 19:40











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1 Answer
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active

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1 Answer
1






active

oldest

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active

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active

oldest

votes








up vote
7
down vote



accepted











Can a concept evaluation depend on where it is evaluated?




Yes. This was explicitly discussed during core wording review when merging Concepts into the working draft. The concept is re-evaluated every time.



As a result, this:



template<class T>
concept Complete = sizeof(T) == sizeof(T);

struct A;
static_assert(!Complete<A>);
struct A {};
static_assert(Complete<A>);


is well-formed. In other words, we don't "memoize" concepts in the same way we "memoize" template instantiations.






share|improve this answer





















  • Thanks! Could you have a look at this question too?
    – Oliv
    Nov 12 at 18:50








  • 1




    Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
    – Davis Herring
    Nov 12 at 19:40















up vote
7
down vote



accepted











Can a concept evaluation depend on where it is evaluated?




Yes. This was explicitly discussed during core wording review when merging Concepts into the working draft. The concept is re-evaluated every time.



As a result, this:



template<class T>
concept Complete = sizeof(T) == sizeof(T);

struct A;
static_assert(!Complete<A>);
struct A {};
static_assert(Complete<A>);


is well-formed. In other words, we don't "memoize" concepts in the same way we "memoize" template instantiations.






share|improve this answer





















  • Thanks! Could you have a look at this question too?
    – Oliv
    Nov 12 at 18:50








  • 1




    Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
    – Davis Herring
    Nov 12 at 19:40













up vote
7
down vote



accepted







up vote
7
down vote



accepted







Can a concept evaluation depend on where it is evaluated?




Yes. This was explicitly discussed during core wording review when merging Concepts into the working draft. The concept is re-evaluated every time.



As a result, this:



template<class T>
concept Complete = sizeof(T) == sizeof(T);

struct A;
static_assert(!Complete<A>);
struct A {};
static_assert(Complete<A>);


is well-formed. In other words, we don't "memoize" concepts in the same way we "memoize" template instantiations.






share|improve this answer













Can a concept evaluation depend on where it is evaluated?




Yes. This was explicitly discussed during core wording review when merging Concepts into the working draft. The concept is re-evaluated every time.



As a result, this:



template<class T>
concept Complete = sizeof(T) == sizeof(T);

struct A;
static_assert(!Complete<A>);
struct A {};
static_assert(Complete<A>);


is well-formed. In other words, we don't "memoize" concepts in the same way we "memoize" template instantiations.







share|improve this answer












share|improve this answer



share|improve this answer










answered Nov 12 at 18:43









Barry

176k18301557




176k18301557












  • Thanks! Could you have a look at this question too?
    – Oliv
    Nov 12 at 18:50








  • 1




    Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
    – Davis Herring
    Nov 12 at 19:40


















  • Thanks! Could you have a look at this question too?
    – Oliv
    Nov 12 at 18:50








  • 1




    Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
    – Davis Herring
    Nov 12 at 19:40
















Thanks! Could you have a look at this question too?
– Oliv
Nov 12 at 18:50






Thanks! Could you have a look at this question too?
– Oliv
Nov 12 at 18:50






1




1




Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
– Davis Herring
Nov 12 at 19:40




Note that it’s ill-formed NDR to have a template whose instantiations might be different because of such changes between two of its (potential) points of instantiation. (Mostly this affects function templates.)
– Davis Herring
Nov 12 at 19:40


















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