Is unsigned int a `simple-type-specifier`

Question

The reason I'm asking this is because, notwithstanding the fact that the title for Table 11 is simple-type-specifiers and the types they specify, the grammar production simple-type-specifier doesn't mention this combination of types, as a simple-type-specifier. So is unsigned int a simple-type-specifier?

Edit: I don't think the answer given to the aforementioned question is correct. If that was true, one would not be able to say that the declaration void* operator new(std::size_t); is a declaration as defined in [dcl.dcl]/1, as size_t is typedef defined (in my system) as unsigned int. To show this assertion, i.e., that void* operator new(std::size_t); is a declaration, I almost sure we need the fact that an unsigned int is a simple-type-specifier.. I was wrong. There is no problem in saying that void* operator new(std::size_t); is a declaration, if we use the definition of type-name given in [dcl.type.simple]/1. I think I was right the first time, except for the wrong example. That is, I believe that unsigned int has to be a simple-type-specifier, otherwise one would not be able to say that void f(unsigned int); is a declaration.

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Answer 1

No, unsigned int is not a simple-type-specifier. If you read the preceding paragraph you have

Table 11 summarizes the valid combinations of simple-type-specifiers and the types they specify.

^{emphasis mine}

So, while both unsigned and int are listed in paragraph 1 as simple-type-specifiers the combination of the two is not, but it is a valid type.