Why doesn't “auto ch = unsigned char{'p'}” compile under C++ 17?












22















I'm puzzled. Isn't const auto ch = unsigned char{'p'}; a perfectly valid initialization expression? Fails to be compiled by all three major compilers with almost identical error messages:




error: expected '(' for function-style cast or type construction




Swapping curly braces for ('p') changes nothing.
It does, however, compile without the signed or unsigned keyword.



Online demo.






c++ c++14 c++17







share|improve this question























  • @FrançoisAndrieux: so does const auto ch = static_cast<unsigned char>('p'), but that's conversion, not initialization.

    – Violet Giraffe
    9 hours ago








  • 3





    using T = unsigned char; const auto ch = T{'p'}; seems to work.

    – François Andrieux
    9 hours ago











  • @FrançoisAndrieux: Hm, do you think the compiler simply fails to parse unsigned char as a single type name in this context?

    – Violet Giraffe
    9 hours ago






  • 1





    const auto ch = (unsigned char){'p'};?

    – Yakk - Adam Nevraumont
    6 hours ago
















22















I'm puzzled. Isn't const auto ch = unsigned char{'p'}; a perfectly valid initialization expression? Fails to be compiled by all three major compilers with almost identical error messages:




error: expected '(' for function-style cast or type construction




Swapping curly braces for ('p') changes nothing.
It does, however, compile without the signed or unsigned keyword.



Online demo.






c++ c++14 c++17







share|improve this question























  • @FrançoisAndrieux: so does const auto ch = static_cast<unsigned char>('p'), but that's conversion, not initialization.

    – Violet Giraffe
    9 hours ago








  • 3





    using T = unsigned char; const auto ch = T{'p'}; seems to work.

    – François Andrieux
    9 hours ago











  • @FrançoisAndrieux: Hm, do you think the compiler simply fails to parse unsigned char as a single type name in this context?

    – Violet Giraffe
    9 hours ago






  • 1





    const auto ch = (unsigned char){'p'};?

    – Yakk - Adam Nevraumont
    6 hours ago














22












22








22


2






I'm puzzled. Isn't const auto ch = unsigned char{'p'}; a perfectly valid initialization expression? Fails to be compiled by all three major compilers with almost identical error messages:




error: expected '(' for function-style cast or type construction




Swapping curly braces for ('p') changes nothing.
It does, however, compile without the signed or unsigned keyword.



Online demo.






c++ c++14 c++17







share|improve this question














I'm puzzled. Isn't const auto ch = unsigned char{'p'}; a perfectly valid initialization expression? Fails to be compiled by all three major compilers with almost identical error messages:




error: expected '(' for function-style cast or type construction




Swapping curly braces for ('p') changes nothing.
It does, however, compile without the signed or unsigned keyword.



Online demo.






c++ c++14 c++17




c++ c++14 c++17






share|improve this question













share|improve this question











share|improve this question




share|improve this question










asked 9 hours ago









Violet GiraffeViolet Giraffe

14.7k28135249




14.7k28135249













  • @FrançoisAndrieux: so does const auto ch = static_cast<unsigned char>('p'), but that's conversion, not initialization.

    – Violet Giraffe
    9 hours ago








  • 3





    using T = unsigned char; const auto ch = T{'p'}; seems to work.

    – François Andrieux
    9 hours ago











  • @FrançoisAndrieux: Hm, do you think the compiler simply fails to parse unsigned char as a single type name in this context?

    – Violet Giraffe
    9 hours ago






  • 1





    const auto ch = (unsigned char){'p'};?

    – Yakk - Adam Nevraumont
    6 hours ago



















  • @FrançoisAndrieux: so does const auto ch = static_cast<unsigned char>('p'), but that's conversion, not initialization.

    – Violet Giraffe
    9 hours ago








  • 3





    using T = unsigned char; const auto ch = T{'p'}; seems to work.

    – François Andrieux
    9 hours ago











  • @FrançoisAndrieux: Hm, do you think the compiler simply fails to parse unsigned char as a single type name in this context?

    – Violet Giraffe
    9 hours ago






  • 1





    const auto ch = (unsigned char){'p'};?

    – Yakk - Adam Nevraumont
    6 hours ago

















@FrançoisAndrieux: so does const auto ch = static_cast<unsigned char>('p'), but that's conversion, not initialization.

– Violet Giraffe
9 hours ago







@FrançoisAndrieux: so does const auto ch = static_cast<unsigned char>('p'), but that's conversion, not initialization.

– Violet Giraffe
9 hours ago






3




3





using T = unsigned char; const auto ch = T{'p'}; seems to work.

– François Andrieux
9 hours ago





using T = unsigned char; const auto ch = T{'p'}; seems to work.

– François Andrieux
9 hours ago













@FrançoisAndrieux: Hm, do you think the compiler simply fails to parse unsigned char as a single type name in this context?

– Violet Giraffe
9 hours ago





@FrançoisAndrieux: Hm, do you think the compiler simply fails to parse unsigned char as a single type name in this context?

– Violet Giraffe
9 hours ago




1




1





const auto ch = (unsigned char){'p'};?

– Yakk - Adam Nevraumont
6 hours ago





const auto ch = (unsigned char){'p'};?

– Yakk - Adam Nevraumont
6 hours ago












1 Answer
1






active

oldest

votes


















31














Because only single-word type name could be used for this kind of explicit type conversion.




A single-word type name followed by a braced-init-list is a prvalue of the specified type designating a temporary (until C++17) whose result object is (since C++17) direct-list-initialized with the specified braced-init-list.




unsigned char is not a single-word type name, while char is. And this is true for functional cast expression too, that's why ('p') doesn't work either.



As the workaround, you can



using uc = unsigned char;  // or use typedef

const auto ch = uc{'p'};


Or change it to other cast styles.



const auto ch = (unsigned char) 'p';  // c-style cast expression

const auto ch = static_cast<unsigned char>('p');  // static_cast conversion





share|improve this answer





















  • 5





    Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

    – Violet Giraffe
    9 hours ago






  • 1





    @VioletGiraffe To be honest I don't know; I only know that the standard says so.

    – songyuanyao
    9 hours ago






  • 1





    @VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

    – NathanOliver
    8 hours ago






  • 1





    @curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

    – NathanOliver
    8 hours ago






  • 4





    C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

    – Arne Vogel
    8 hours ago











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31














Because only single-word type name could be used for this kind of explicit type conversion.




A single-word type name followed by a braced-init-list is a prvalue of the specified type designating a temporary (until C++17) whose result object is (since C++17) direct-list-initialized with the specified braced-init-list.




unsigned char is not a single-word type name, while char is. And this is true for functional cast expression too, that's why ('p') doesn't work either.



As the workaround, you can



using uc = unsigned char;  // or use typedef

const auto ch = uc{'p'};


Or change it to other cast styles.



const auto ch = (unsigned char) 'p';  // c-style cast expression

const auto ch = static_cast<unsigned char>('p');  // static_cast conversion





share|improve this answer





















  • 5





    Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

    – Violet Giraffe
    9 hours ago






  • 1





    @VioletGiraffe To be honest I don't know; I only know that the standard says so.

    – songyuanyao
    9 hours ago






  • 1





    @VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

    – NathanOliver
    8 hours ago






  • 1





    @curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

    – NathanOliver
    8 hours ago






  • 4





    C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

    – Arne Vogel
    8 hours ago
















31














Because only single-word type name could be used for this kind of explicit type conversion.




A single-word type name followed by a braced-init-list is a prvalue of the specified type designating a temporary (until C++17) whose result object is (since C++17) direct-list-initialized with the specified braced-init-list.




unsigned char is not a single-word type name, while char is. And this is true for functional cast expression too, that's why ('p') doesn't work either.



As the workaround, you can



using uc = unsigned char;  // or use typedef

const auto ch = uc{'p'};


Or change it to other cast styles.



const auto ch = (unsigned char) 'p';  // c-style cast expression

const auto ch = static_cast<unsigned char>('p');  // static_cast conversion





share|improve this answer





















  • 5





    Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

    – Violet Giraffe
    9 hours ago






  • 1





    @VioletGiraffe To be honest I don't know; I only know that the standard says so.

    – songyuanyao
    9 hours ago






  • 1





    @VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

    – NathanOliver
    8 hours ago






  • 1





    @curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

    – NathanOliver
    8 hours ago






  • 4





    C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

    – Arne Vogel
    8 hours ago














31












31








31







Because only single-word type name could be used for this kind of explicit type conversion.




A single-word type name followed by a braced-init-list is a prvalue of the specified type designating a temporary (until C++17) whose result object is (since C++17) direct-list-initialized with the specified braced-init-list.




unsigned char is not a single-word type name, while char is. And this is true for functional cast expression too, that's why ('p') doesn't work either.



As the workaround, you can



using uc = unsigned char;  // or use typedef

const auto ch = uc{'p'};


Or change it to other cast styles.



const auto ch = (unsigned char) 'p';  // c-style cast expression

const auto ch = static_cast<unsigned char>('p');  // static_cast conversion





share|improve this answer















Because only single-word type name could be used for this kind of explicit type conversion.




A single-word type name followed by a braced-init-list is a prvalue of the specified type designating a temporary (until C++17) whose result object is (since C++17) direct-list-initialized with the specified braced-init-list.




unsigned char is not a single-word type name, while char is. And this is true for functional cast expression too, that's why ('p') doesn't work either.



As the workaround, you can



using uc = unsigned char;  // or use typedef

const auto ch = uc{'p'};


Or change it to other cast styles.



const auto ch = (unsigned char) 'p';  // c-style cast expression

const auto ch = static_cast<unsigned char>('p');  // static_cast conversion






share|improve this answer














share|improve this answer



share|improve this answer








edited 8 hours ago

























answered 9 hours ago









songyuanyaosongyuanyao

92.2k11177242




92.2k11177242








  • 5





    Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

    – Violet Giraffe
    9 hours ago






  • 1





    @VioletGiraffe To be honest I don't know; I only know that the standard says so.

    – songyuanyao
    9 hours ago






  • 1





    @VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

    – NathanOliver
    8 hours ago






  • 1





    @curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

    – NathanOliver
    8 hours ago






  • 4





    C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

    – Arne Vogel
    8 hours ago














  • 5





    Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

    – Violet Giraffe
    9 hours ago






  • 1





    @VioletGiraffe To be honest I don't know; I only know that the standard says so.

    – songyuanyao
    9 hours ago






  • 1





    @VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

    – NathanOliver
    8 hours ago






  • 1





    @curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

    – NathanOliver
    8 hours ago






  • 4





    C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

    – Arne Vogel
    8 hours ago








5




5





Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

– Violet Giraffe
9 hours ago





Do you happen to know the reason for this limitation? Like, if multi-word type names were allowed here, what other things would become broken?

– Violet Giraffe
9 hours ago




1




1





@VioletGiraffe To be honest I don't know; I only know that the standard says so.

– songyuanyao
9 hours ago





@VioletGiraffe To be honest I don't know; I only know that the standard says so.

– songyuanyao
9 hours ago




1




1





@VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

– NathanOliver
8 hours ago





@VioletGiraffe Because uc{'p'}/uc('p') is a functional cast. A function name can't have a space in it so neither can the type name.

– NathanOliver
8 hours ago




1




1





@curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

– NathanOliver
8 hours ago





@curiousguy For all intents and purposes it is. If the type is a class type then the constructor is called. the cast and call have the exact same syntax and grammar.

– NathanOliver
8 hours ago




4




4





C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

– Arne Vogel
8 hours ago





C-style cast, really? How about static_cast instead (or in case Boost is used, boost::implicit_cast)? The language allows a simple-type-specifier or (in a template) a typename-specifier, BTW, in a function-style cast, to be precise. Your statement that it "can't have a space" is slightly misleading because e.g. std :: uint8_t has spaces and is a valid simple-type-specifier.

– Arne Vogel
8 hours ago




















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