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

What is an unsigned char?Why does C++ compilation take so long?Why do we need virtual functions in C++?Why is this program erroneously rejected by three C++ compilers?Why should C++ programmers minimize use of 'new'?Why is reading lines from stdin much slower in C++ than Python?g++ gives error : invalid initialization of reference of type ‘char&’ from expression of type ‘unsigned char’basic_string of unsigned char Value TypeError when attempting to static_cast<> const char[] to unsigned char *How do I cast a const char* to a const unsigned char*

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

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  using T = unsigned char; const auto ch = T{'p'}; seems to work.
  
  – François Andrieux
  15 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
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  const auto ch = (unsigned char){'p'};?
  
  – Yakk - Adam Nevraumont
  11 hours ago
  

  
  
  
  

add a comment | 

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

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

• 
  
  
  

  
  

  
  
  
  
  
  

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

  
  
  
  


• 
  
  
  

  
  3
  

  
  
  
  
  
  

  
  
  using T = unsigned char; const auto ch = T{'p'}; seems to work.
  
  – François Andrieux
  15 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
  15 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  const auto ch = (unsigned char){'p'};?
  
  – Yakk - Adam Nevraumont
  11 hours ago
  

  
  
  
  

add a comment | 

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

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

share|improve this question

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

share|improve this question

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

asked 15 hours ago

Violet GiraffeViolet Giraffe

14.7k28135249

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

edited 13 hours ago

answered 14 hours ago

songyuanyaosongyuanyao

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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @VioletGiraffe To be honest I don't know; I only know that the standard says so.
  
  – songyuanyao
  14 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
  13 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
  13 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
  13 hours ago
  

  
  
  
  

 | 
show 2 more comments

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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

edited 13 hours ago

answered 14 hours ago

songyuanyaosongyuanyao

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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @VioletGiraffe To be honest I don't know; I only know that the standard says so.
  
  – songyuanyao
  14 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
  13 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
  13 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
  13 hours ago
  

  
  
  
  

 | 
show 2 more comments

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

edited 13 hours ago

answered 14 hours ago

songyuanyaosongyuanyao

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
  14 hours ago
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  
  @VioletGiraffe To be honest I don't know; I only know that the standard says so.
  
  – songyuanyao
  14 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
  13 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
  13 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
  13 hours ago
  

  
  
  
  

 | 
show 2 more comments

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

edited 13 hours ago

answered 14 hours ago

songyuanyaosongyuanyao

92.2k11177242

using uc = unsigned char;  // or use typedef

const auto ch = uc{'p'};

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

edited 13 hours ago

answered 14 hours ago

songyuanyaosongyuanyao

92.2k11177242

answered 14 hours ago

songyuanyaosongyuanyao

92.2k11177242

answered 14 hours ago

songyuanyaosongyuanyao

92.2k11177242