Why doesn't unsigned typeof(var) work? const typeof(x) works fine, so why not?

Question

In a macro I would want to make the value the same width, however unsigned. So, a natural unsigned typeof(x) came to mind. However, it doesn't compile:

/home/guest/mc.c: In function ‘main’:
/home/guest/mc.c:14:36: error: expected ‘)’ before ‘typeof’
   14 |                         *((unsigned typeof(minus)*)&minus));

I wonder why it is so? Maybe I'm doing something wrong? Can I somehow add unsigned to the type of a variable?

PS. I've noted, that similar cast adding const works OK:

#define add_const(x) ((const typeof(x))(x))

Also, adding * to create a pointer type also works OK.

Answer 1

The fundamental reason why unsigned typeof(x) doesn't work but const typeof(x) does is that unsigned and const are two different syntax components. unsigned it a type specifier, while const is a type qualifier.

The list of type specifiers (note that a bare unsigned is syntactically equivalent to unsigned int:

6.7.2 Type specifiers

1. Syntax

      type-specifier:
             void
             char
             short
             int
             long
             float
             double
             signed
             unsigned
             _Bool
             _Complex
             atomic-type-specifier
             struct-or-union-specifier
             enum-specifier
             typedef-name

2. Constraints

...

• unsigned, or unsigned int

So unsigned typeof(x) is the equivalent of unsigned int typeof(x).

const is a type qualifier:

6.7.3 Type qualifiers

1. Syntax

      type-qualifier:
             const
             restrict
             volatile
             _Atomic

const char is proper C code, assuming x is a char. unsigned int char is not proper C code.