Is unsigned int `simple-type-specifier`
The reason I am asking this is because, even though the title for Table 11 is a simple type specifier, and the types they specify, the release of the grammar simple type specifier does not mention this combination of types as a specifier simple type. So unsigned int
a simple type specifier?
Edit : I don't think the answer to the above question is correct. If this were the case, it would not be possible to say that the declaration ... void* operator new(std::size_t);
is a declaration as defined in [dcl.dcl] / 1 , since it size_t
is defined typedef
(on my system) as unsigned int
. To show this statement, that is, what void* operator new(std::size_t);
is a declaration, I'm pretty sure we need to unsigned int
be a simple type specifier.I was wrong. There is no point in saying what void* operator new(std::size_t);
is a declaration if we use the type-type definition given in [dcl.type.simple] / 1hit>. I think I was right the first time, except for the wrong example. That is, I believe that there unsigned int
should be a simple type-specifier , otherwise it would not be possible to say what it void f(unsigned int);
is.
...
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No, unsigned int
it is not a simple type specifier. If you have read the previous paragraph , you
Table 11 lists the allowed combinations of simple type specifiers and the types they define.
my accent
So, although both unsigned
and int
are listed in paragraph 1 as a simple type specifiers, the combination of these two do not have, but it is a valid type.
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