Enforcing function contract at compile time when possible










13














(this question was inspired by How can I generate a compilation error to prevent certain VALUE (not type) to go into the function?)



Let's say, we have a single-argument foo, semantically defined as



int foo(int arg) 
int* parg;
if (arg != 5)
parg = &arg;


return *parg;



The whole code above is used to illustrate a simple idea - function returns it's own argument unless the argument is equal to 5, in which case behavior is undefined.



Now, the challenge - modify the function in such a way, that if it's argument is known at compile time, a compiler diagnostic (warning or error) should be generated, and if not, behavior remains undefined in runtime. Solution could be compiler-dependent, as long as it is available in either one of the big 4 compilers.



Here are some potential routes which do not solve the problem:



  • Making function a template which takes it's argument as a template parameter - this doesn't solve the problem because it makes function ineligible for run-time arguments

  • Making function a constexpr - this doesn't solve the problem, because even when compilers see undefined behavior, they do not produce diagnostics in my tests - instead, gcc inserts ud2 instruction, which is not what I want.









share|improve this question



















  • 6




    out of curiosity, why this Q is upvoted while linked-one is downvoted?
    – apple apple
    yesterday






  • 1




    @appleapple Because this one is #1 well-formulated, #2 gives a short a suffisant context and #3 defines a precise (SMART) objective.
    – YSC
    yesterday






  • 6




    @YSC well, this question does not show any research effort; it's unclear or not useful. I'd give a downvote if I have to cast one.
    – apple apple
    yesterday






  • 1




    @appleapple You don't have to cast one, but if you want to, feel free to do so. This is a matter of taste I guess. I find it useful (I never succeded at that task and I think it would be nice to have an API making a value-related contract enforced by a compiler error).
    – YSC
    yesterday






  • 3




    @IłyaBursov - not a duplicate, suggested answer doesn't solve the problem.
    – SergeyA
    yesterday















13














(this question was inspired by How can I generate a compilation error to prevent certain VALUE (not type) to go into the function?)



Let's say, we have a single-argument foo, semantically defined as



int foo(int arg) 
int* parg;
if (arg != 5)
parg = &arg;


return *parg;



The whole code above is used to illustrate a simple idea - function returns it's own argument unless the argument is equal to 5, in which case behavior is undefined.



Now, the challenge - modify the function in such a way, that if it's argument is known at compile time, a compiler diagnostic (warning or error) should be generated, and if not, behavior remains undefined in runtime. Solution could be compiler-dependent, as long as it is available in either one of the big 4 compilers.



Here are some potential routes which do not solve the problem:



  • Making function a template which takes it's argument as a template parameter - this doesn't solve the problem because it makes function ineligible for run-time arguments

  • Making function a constexpr - this doesn't solve the problem, because even when compilers see undefined behavior, they do not produce diagnostics in my tests - instead, gcc inserts ud2 instruction, which is not what I want.









share|improve this question



















  • 6




    out of curiosity, why this Q is upvoted while linked-one is downvoted?
    – apple apple
    yesterday






  • 1




    @appleapple Because this one is #1 well-formulated, #2 gives a short a suffisant context and #3 defines a precise (SMART) objective.
    – YSC
    yesterday






  • 6




    @YSC well, this question does not show any research effort; it's unclear or not useful. I'd give a downvote if I have to cast one.
    – apple apple
    yesterday






  • 1




    @appleapple You don't have to cast one, but if you want to, feel free to do so. This is a matter of taste I guess. I find it useful (I never succeded at that task and I think it would be nice to have an API making a value-related contract enforced by a compiler error).
    – YSC
    yesterday






  • 3




    @IłyaBursov - not a duplicate, suggested answer doesn't solve the problem.
    – SergeyA
    yesterday













13












13








13


6





(this question was inspired by How can I generate a compilation error to prevent certain VALUE (not type) to go into the function?)



Let's say, we have a single-argument foo, semantically defined as



int foo(int arg) 
int* parg;
if (arg != 5)
parg = &arg;


return *parg;



The whole code above is used to illustrate a simple idea - function returns it's own argument unless the argument is equal to 5, in which case behavior is undefined.



Now, the challenge - modify the function in such a way, that if it's argument is known at compile time, a compiler diagnostic (warning or error) should be generated, and if not, behavior remains undefined in runtime. Solution could be compiler-dependent, as long as it is available in either one of the big 4 compilers.



Here are some potential routes which do not solve the problem:



  • Making function a template which takes it's argument as a template parameter - this doesn't solve the problem because it makes function ineligible for run-time arguments

  • Making function a constexpr - this doesn't solve the problem, because even when compilers see undefined behavior, they do not produce diagnostics in my tests - instead, gcc inserts ud2 instruction, which is not what I want.









share|improve this question















(this question was inspired by How can I generate a compilation error to prevent certain VALUE (not type) to go into the function?)



Let's say, we have a single-argument foo, semantically defined as



int foo(int arg) 
int* parg;
if (arg != 5)
parg = &arg;


return *parg;



The whole code above is used to illustrate a simple idea - function returns it's own argument unless the argument is equal to 5, in which case behavior is undefined.



Now, the challenge - modify the function in such a way, that if it's argument is known at compile time, a compiler diagnostic (warning or error) should be generated, and if not, behavior remains undefined in runtime. Solution could be compiler-dependent, as long as it is available in either one of the big 4 compilers.



Here are some potential routes which do not solve the problem:



  • Making function a template which takes it's argument as a template parameter - this doesn't solve the problem because it makes function ineligible for run-time arguments

  • Making function a constexpr - this doesn't solve the problem, because even when compilers see undefined behavior, they do not produce diagnostics in my tests - instead, gcc inserts ud2 instruction, which is not what I want.






c++






share|improve this question















share|improve this question













share|improve this question




share|improve this question








edited yesterday

























asked yesterday









SergeyA

41.3k53783




41.3k53783







  • 6




    out of curiosity, why this Q is upvoted while linked-one is downvoted?
    – apple apple
    yesterday






  • 1




    @appleapple Because this one is #1 well-formulated, #2 gives a short a suffisant context and #3 defines a precise (SMART) objective.
    – YSC
    yesterday






  • 6




    @YSC well, this question does not show any research effort; it's unclear or not useful. I'd give a downvote if I have to cast one.
    – apple apple
    yesterday






  • 1




    @appleapple You don't have to cast one, but if you want to, feel free to do so. This is a matter of taste I guess. I find it useful (I never succeded at that task and I think it would be nice to have an API making a value-related contract enforced by a compiler error).
    – YSC
    yesterday






  • 3




    @IłyaBursov - not a duplicate, suggested answer doesn't solve the problem.
    – SergeyA
    yesterday












  • 6




    out of curiosity, why this Q is upvoted while linked-one is downvoted?
    – apple apple
    yesterday






  • 1




    @appleapple Because this one is #1 well-formulated, #2 gives a short a suffisant context and #3 defines a precise (SMART) objective.
    – YSC
    yesterday






  • 6




    @YSC well, this question does not show any research effort; it's unclear or not useful. I'd give a downvote if I have to cast one.
    – apple apple
    yesterday






  • 1




    @appleapple You don't have to cast one, but if you want to, feel free to do so. This is a matter of taste I guess. I find it useful (I never succeded at that task and I think it would be nice to have an API making a value-related contract enforced by a compiler error).
    – YSC
    yesterday






  • 3




    @IłyaBursov - not a duplicate, suggested answer doesn't solve the problem.
    – SergeyA
    yesterday







6




6




out of curiosity, why this Q is upvoted while linked-one is downvoted?
– apple apple
yesterday




out of curiosity, why this Q is upvoted while linked-one is downvoted?
– apple apple
yesterday




1




1




@appleapple Because this one is #1 well-formulated, #2 gives a short a suffisant context and #3 defines a precise (SMART) objective.
– YSC
yesterday




@appleapple Because this one is #1 well-formulated, #2 gives a short a suffisant context and #3 defines a precise (SMART) objective.
– YSC
yesterday




6




6




@YSC well, this question does not show any research effort; it's unclear or not useful. I'd give a downvote if I have to cast one.
– apple apple
yesterday




@YSC well, this question does not show any research effort; it's unclear or not useful. I'd give a downvote if I have to cast one.
– apple apple
yesterday




1




1




@appleapple You don't have to cast one, but if you want to, feel free to do so. This is a matter of taste I guess. I find it useful (I never succeded at that task and I think it would be nice to have an API making a value-related contract enforced by a compiler error).
– YSC
yesterday




@appleapple You don't have to cast one, but if you want to, feel free to do so. This is a matter of taste I guess. I find it useful (I never succeded at that task and I think it would be nice to have an API making a value-related contract enforced by a compiler error).
– YSC
yesterday




3




3




@IłyaBursov - not a duplicate, suggested answer doesn't solve the problem.
– SergeyA
yesterday




@IłyaBursov - not a duplicate, suggested answer doesn't solve the problem.
– SergeyA
yesterday












3 Answers
3






active

oldest

votes


















4














I got error with constexpr when used in constant expression for:



constexpr int foo(int arg) 
int* parg = nullptr;
if (arg != 5)
parg = &arg;

return *parg;



Demo



We cannot know that argument value is known at compile type, but we can use type representing value with std::integral_constant



// alias to shorten name. 
template <int N>
using int_c = std::integral_constant<int, N>;


Possibly with UDL with operator "" _c to have 5_c, 42_c.



and then, add overload with that:



template <int N>
constexpr auto foo(int_c<N>)
return int_c<foo(N)>;



So:



foo(int_c<42>); // OK
foo(int_c<5>); // Fail to compile

// and with previous constexpr:
foo(5); // Runtime error, No compile time diagnostic
constexpr auto r = foo(5); // Fail to compile


As I said, arguments are not known to be constant inside the function, and is_constexpr seems not possible in standard to allow dispatch, but some compiler provide built-in for that (__builtin_constant_p), so with MACRO, we can do the dispatch:



#define FOO(X) [&]() 
if constexpr (__builtin_constant_p(X))
return foo(int_c<__builtin_constant_p (X) ? X : 0>);
else
return foo(X);

()


Demo



Note: Cannot use foo(int_c<X>) directly, even in if constexpr, as there is still some syntax check.






share|improve this answer






















  • Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
    – SergeyA
    yesterday











  • I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
    – SergeyA
    yesterday











  • That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
    – Jarod42
    yesterday











  • Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
    – SergeyA
    yesterday










  • Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
    – Jarod42
    yesterday


















2














gcc/clang/intel compilers support __builtin_constant_p, so you can use something like that:



template <int D>
int foo_ub(int arg)
static_assert(D != 5, "error");
int* parg = nullptr;
if (arg != 5)
parg = &arg;


return *parg;


#define foo(e) foo_ub< __builtin_constant_p(e) ? e : 0 >(e)


these statements produce compile time error:



  • foo(5)

  • foo(2+3)

  • constexpr int i = 5; foo(i);

while all others - runtime segfault (or ub if no nullptr is used)






share|improve this answer




























    0














    It's not perfect and it requires us to use arguments in two different places, but it 'works':



    template<int N = 0>
    int foo(int arg = 0)
    static_assert(N != 5, "N cannot be 5!");
    int* parg;
    if (arg != 5)
    parg = &arg;


    return *parg;



    We can call it like so:



    foo<5>(); // does not compile
    foo(5); // UB
    foo<5>(5); // does not compile
    foo<5>(10); // does not compile
    foo<10>(5); // UB
    foo(); // fine
    foo<10>(); // fine
    foo(10); // fine





    share|improve this answer
















    • 2




      No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
      – SergeyA
      yesterday






    • 1




      Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
      – Fureeish
      yesterday










    • You might add an additional runtime check that argument are equal or defaulted.
      – Jarod42
      yesterday










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    3 Answers
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    active

    oldest

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    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4














    I got error with constexpr when used in constant expression for:



    constexpr int foo(int arg) 
    int* parg = nullptr;
    if (arg != 5)
    parg = &arg;

    return *parg;



    Demo



    We cannot know that argument value is known at compile type, but we can use type representing value with std::integral_constant



    // alias to shorten name. 
    template <int N>
    using int_c = std::integral_constant<int, N>;


    Possibly with UDL with operator "" _c to have 5_c, 42_c.



    and then, add overload with that:



    template <int N>
    constexpr auto foo(int_c<N>)
    return int_c<foo(N)>;



    So:



    foo(int_c<42>); // OK
    foo(int_c<5>); // Fail to compile

    // and with previous constexpr:
    foo(5); // Runtime error, No compile time diagnostic
    constexpr auto r = foo(5); // Fail to compile


    As I said, arguments are not known to be constant inside the function, and is_constexpr seems not possible in standard to allow dispatch, but some compiler provide built-in for that (__builtin_constant_p), so with MACRO, we can do the dispatch:



    #define FOO(X) [&]() 
    if constexpr (__builtin_constant_p(X))
    return foo(int_c<__builtin_constant_p (X) ? X : 0>);
    else
    return foo(X);

    ()


    Demo



    Note: Cannot use foo(int_c<X>) directly, even in if constexpr, as there is still some syntax check.






    share|improve this answer






















    • Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
      – SergeyA
      yesterday











    • I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
      – SergeyA
      yesterday











    • That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
      – Jarod42
      yesterday











    • Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
      – SergeyA
      yesterday










    • Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
      – Jarod42
      yesterday















    4














    I got error with constexpr when used in constant expression for:



    constexpr int foo(int arg) 
    int* parg = nullptr;
    if (arg != 5)
    parg = &arg;

    return *parg;



    Demo



    We cannot know that argument value is known at compile type, but we can use type representing value with std::integral_constant



    // alias to shorten name. 
    template <int N>
    using int_c = std::integral_constant<int, N>;


    Possibly with UDL with operator "" _c to have 5_c, 42_c.



    and then, add overload with that:



    template <int N>
    constexpr auto foo(int_c<N>)
    return int_c<foo(N)>;



    So:



    foo(int_c<42>); // OK
    foo(int_c<5>); // Fail to compile

    // and with previous constexpr:
    foo(5); // Runtime error, No compile time diagnostic
    constexpr auto r = foo(5); // Fail to compile


    As I said, arguments are not known to be constant inside the function, and is_constexpr seems not possible in standard to allow dispatch, but some compiler provide built-in for that (__builtin_constant_p), so with MACRO, we can do the dispatch:



    #define FOO(X) [&]() 
    if constexpr (__builtin_constant_p(X))
    return foo(int_c<__builtin_constant_p (X) ? X : 0>);
    else
    return foo(X);

    ()


    Demo



    Note: Cannot use foo(int_c<X>) directly, even in if constexpr, as there is still some syntax check.






    share|improve this answer






















    • Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
      – SergeyA
      yesterday











    • I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
      – SergeyA
      yesterday











    • That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
      – Jarod42
      yesterday











    • Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
      – SergeyA
      yesterday










    • Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
      – Jarod42
      yesterday













    4












    4








    4






    I got error with constexpr when used in constant expression for:



    constexpr int foo(int arg) 
    int* parg = nullptr;
    if (arg != 5)
    parg = &arg;

    return *parg;



    Demo



    We cannot know that argument value is known at compile type, but we can use type representing value with std::integral_constant



    // alias to shorten name. 
    template <int N>
    using int_c = std::integral_constant<int, N>;


    Possibly with UDL with operator "" _c to have 5_c, 42_c.



    and then, add overload with that:



    template <int N>
    constexpr auto foo(int_c<N>)
    return int_c<foo(N)>;



    So:



    foo(int_c<42>); // OK
    foo(int_c<5>); // Fail to compile

    // and with previous constexpr:
    foo(5); // Runtime error, No compile time diagnostic
    constexpr auto r = foo(5); // Fail to compile


    As I said, arguments are not known to be constant inside the function, and is_constexpr seems not possible in standard to allow dispatch, but some compiler provide built-in for that (__builtin_constant_p), so with MACRO, we can do the dispatch:



    #define FOO(X) [&]() 
    if constexpr (__builtin_constant_p(X))
    return foo(int_c<__builtin_constant_p (X) ? X : 0>);
    else
    return foo(X);

    ()


    Demo



    Note: Cannot use foo(int_c<X>) directly, even in if constexpr, as there is still some syntax check.






    share|improve this answer














    I got error with constexpr when used in constant expression for:



    constexpr int foo(int arg) 
    int* parg = nullptr;
    if (arg != 5)
    parg = &arg;

    return *parg;



    Demo



    We cannot know that argument value is known at compile type, but we can use type representing value with std::integral_constant



    // alias to shorten name. 
    template <int N>
    using int_c = std::integral_constant<int, N>;


    Possibly with UDL with operator "" _c to have 5_c, 42_c.



    and then, add overload with that:



    template <int N>
    constexpr auto foo(int_c<N>)
    return int_c<foo(N)>;



    So:



    foo(int_c<42>); // OK
    foo(int_c<5>); // Fail to compile

    // and with previous constexpr:
    foo(5); // Runtime error, No compile time diagnostic
    constexpr auto r = foo(5); // Fail to compile


    As I said, arguments are not known to be constant inside the function, and is_constexpr seems not possible in standard to allow dispatch, but some compiler provide built-in for that (__builtin_constant_p), so with MACRO, we can do the dispatch:



    #define FOO(X) [&]() 
    if constexpr (__builtin_constant_p(X))
    return foo(int_c<__builtin_constant_p (X) ? X : 0>);
    else
    return foo(X);

    ()


    Demo



    Note: Cannot use foo(int_c<X>) directly, even in if constexpr, as there is still some syntax check.







    share|improve this answer














    share|improve this answer



    share|improve this answer








    edited 14 hours ago

























    answered yesterday









    Jarod42

    112k1299179




    112k1299179











    • Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
      – SergeyA
      yesterday











    • I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
      – SergeyA
      yesterday











    • That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
      – Jarod42
      yesterday











    • Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
      – SergeyA
      yesterday










    • Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
      – Jarod42
      yesterday
















    • Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
      – SergeyA
      yesterday











    • I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
      – SergeyA
      yesterday











    • That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
      – Jarod42
      yesterday











    • Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
      – SergeyA
      yesterday










    • Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
      – Jarod42
      yesterday















    Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
    – SergeyA
    yesterday





    Constexpr version doesn't give me any diagnostic gcc.godbolt.org/z/ZCho3b when it's result is not used to initialize constant expression variable.
    – SergeyA
    yesterday













    I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
    – SergeyA
    yesterday





    I do understand that, and this is exactly what I am referring to - unless the function is called in constant expression, no diagnostics are provided
    – SergeyA
    yesterday













    That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
    – Jarod42
    yesterday





    That's why I provide in my answer way to pass compile time argument (with int_c) which allows to check at compile time , as they call it in constant expression.
    – Jarod42
    yesterday













    Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
    – SergeyA
    yesterday




    Jarod, may be I am not making myself clear enough. The goal is to have a callable function, which would enforce contract in compile time when it possible (i.e. argument is known). Your solution effectively introduces two overloads, and successful compile-time check is predicated on developers discipline to call second overload.
    – SergeyA
    yesterday












    Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
    – Jarod42
    yesterday




    Found a way with built-in of gcc supported by clang (and so works on clang but not with g++ ;-) )
    – Jarod42
    yesterday













    2














    gcc/clang/intel compilers support __builtin_constant_p, so you can use something like that:



    template <int D>
    int foo_ub(int arg)
    static_assert(D != 5, "error");
    int* parg = nullptr;
    if (arg != 5)
    parg = &arg;


    return *parg;


    #define foo(e) foo_ub< __builtin_constant_p(e) ? e : 0 >(e)


    these statements produce compile time error:



    • foo(5)

    • foo(2+3)

    • constexpr int i = 5; foo(i);

    while all others - runtime segfault (or ub if no nullptr is used)






    share|improve this answer

























      2














      gcc/clang/intel compilers support __builtin_constant_p, so you can use something like that:



      template <int D>
      int foo_ub(int arg)
      static_assert(D != 5, "error");
      int* parg = nullptr;
      if (arg != 5)
      parg = &arg;


      return *parg;


      #define foo(e) foo_ub< __builtin_constant_p(e) ? e : 0 >(e)


      these statements produce compile time error:



      • foo(5)

      • foo(2+3)

      • constexpr int i = 5; foo(i);

      while all others - runtime segfault (or ub if no nullptr is used)






      share|improve this answer























        2












        2








        2






        gcc/clang/intel compilers support __builtin_constant_p, so you can use something like that:



        template <int D>
        int foo_ub(int arg)
        static_assert(D != 5, "error");
        int* parg = nullptr;
        if (arg != 5)
        parg = &arg;


        return *parg;


        #define foo(e) foo_ub< __builtin_constant_p(e) ? e : 0 >(e)


        these statements produce compile time error:



        • foo(5)

        • foo(2+3)

        • constexpr int i = 5; foo(i);

        while all others - runtime segfault (or ub if no nullptr is used)






        share|improve this answer












        gcc/clang/intel compilers support __builtin_constant_p, so you can use something like that:



        template <int D>
        int foo_ub(int arg)
        static_assert(D != 5, "error");
        int* parg = nullptr;
        if (arg != 5)
        parg = &arg;


        return *parg;


        #define foo(e) foo_ub< __builtin_constant_p(e) ? e : 0 >(e)


        these statements produce compile time error:



        • foo(5)

        • foo(2+3)

        • constexpr int i = 5; foo(i);

        while all others - runtime segfault (or ub if no nullptr is used)







        share|improve this answer












        share|improve this answer



        share|improve this answer










        answered yesterday









        Iłya Bursov

        17.9k32543




        17.9k32543





















            0














            It's not perfect and it requires us to use arguments in two different places, but it 'works':



            template<int N = 0>
            int foo(int arg = 0)
            static_assert(N != 5, "N cannot be 5!");
            int* parg;
            if (arg != 5)
            parg = &arg;


            return *parg;



            We can call it like so:



            foo<5>(); // does not compile
            foo(5); // UB
            foo<5>(5); // does not compile
            foo<5>(10); // does not compile
            foo<10>(5); // UB
            foo(); // fine
            foo<10>(); // fine
            foo(10); // fine





            share|improve this answer
















            • 2




              No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
              – SergeyA
              yesterday






            • 1




              Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
              – Fureeish
              yesterday










            • You might add an additional runtime check that argument are equal or defaulted.
              – Jarod42
              yesterday















            0














            It's not perfect and it requires us to use arguments in two different places, but it 'works':



            template<int N = 0>
            int foo(int arg = 0)
            static_assert(N != 5, "N cannot be 5!");
            int* parg;
            if (arg != 5)
            parg = &arg;


            return *parg;



            We can call it like so:



            foo<5>(); // does not compile
            foo(5); // UB
            foo<5>(5); // does not compile
            foo<5>(10); // does not compile
            foo<10>(5); // UB
            foo(); // fine
            foo<10>(); // fine
            foo(10); // fine





            share|improve this answer
















            • 2




              No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
              – SergeyA
              yesterday






            • 1




              Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
              – Fureeish
              yesterday










            • You might add an additional runtime check that argument are equal or defaulted.
              – Jarod42
              yesterday













            0












            0








            0






            It's not perfect and it requires us to use arguments in two different places, but it 'works':



            template<int N = 0>
            int foo(int arg = 0)
            static_assert(N != 5, "N cannot be 5!");
            int* parg;
            if (arg != 5)
            parg = &arg;


            return *parg;



            We can call it like so:



            foo<5>(); // does not compile
            foo(5); // UB
            foo<5>(5); // does not compile
            foo<5>(10); // does not compile
            foo<10>(5); // UB
            foo(); // fine
            foo<10>(); // fine
            foo(10); // fine





            share|improve this answer












            It's not perfect and it requires us to use arguments in two different places, but it 'works':



            template<int N = 0>
            int foo(int arg = 0)
            static_assert(N != 5, "N cannot be 5!");
            int* parg;
            if (arg != 5)
            parg = &arg;


            return *parg;



            We can call it like so:



            foo<5>(); // does not compile
            foo(5); // UB
            foo<5>(5); // does not compile
            foo<5>(10); // does not compile
            foo<10>(5); // UB
            foo(); // fine
            foo<10>(); // fine
            foo(10); // fine






            share|improve this answer












            share|improve this answer



            share|improve this answer










            answered yesterday









            Fureeish

            3,00521025




            3,00521025







            • 2




              No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
              – SergeyA
              yesterday






            • 1




              Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
              – Fureeish
              yesterday










            • You might add an additional runtime check that argument are equal or defaulted.
              – Jarod42
              yesterday












            • 2




              No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
              – SergeyA
              yesterday






            • 1




              Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
              – Fureeish
              yesterday










            • You might add an additional runtime check that argument are equal or defaulted.
              – Jarod42
              yesterday







            2




            2




            No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
            – SergeyA
            yesterday




            No, it doesn't work. Behavior is dependent on programmer's discipline (making sure to provide template argument when it's known at compile time). Instead of this approach, if programmer is disciplined, I'd simply have two functions - templated and not.
            – SergeyA
            yesterday




            1




            1




            Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
            – Fureeish
            yesterday




            Fair, thank you for the comment. I totally agree that this does not actually solve the entire problem, but I will leave it as a neutral hint / information for future visitors :>
            – Fureeish
            yesterday












            You might add an additional runtime check that argument are equal or defaulted.
            – Jarod42
            yesterday




            You might add an additional runtime check that argument are equal or defaulted.
            – Jarod42
            yesterday

















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