Is `unsigned int` a simple-type-specifier?

[Alexander]

According to this answer, to a question of mine in SO, an unsigned int is not a simple-type-specifier. My argument is that it must be a simple-type-specifier, otherwise void f(unsigned int); would not be considered a declaration given the C++ grammar. For instance, if unsigned int is not a simple-type-specifier, it's not a type-specifier, not a defining-type-specifier,  not a decl-specifier, not a decl-specifier-seq, not a parameter-declaration, not a parameter-declaration-list, and so an (unsigned int)  would not be a parameters-and-qualifiers. Continuing this way we would conclude that void f(unsigned int); is would not be a declaration, which is clearly incorrect. What am I missing here?

[Brian Bi]

On Fri, May 26, 2017 at 2:58 PM, Alexander <alexandreter...@gmail.com> wrote:

According to this answer, to a question of mine in SO, an unsigned int is not a simple-type-specifier. My argument is that it must be a simple-type-specifier, otherwise void f(unsigned int); would not be considered a declaration given the C++ grammar. For instance, if unsigned int is not a simple-type-specifier, it's not a type-specifier, not a defining-type-specifier,  not a decl-specifier, not a decl-specifier-seq,

unsigned int is a decl-specifier-seq.

 

not a parameter-declaration, not a parameter-declaration-list, and so an (unsigned int)  would not be a parameters-and-qualifiers. Continuing this way we would conclude that void f(unsigned int); is would not be a declaration, which is clearly incorrect. What am I missing here?

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Brian Bi

[Alexander]

On Friday, May 26, 2017 at 7:07:40 PM UTC-3, Brian Bi wrote:

unsigned int is a decl-specifier-seq.

Great! Thanks.