Why is char neither signed or unsigned, but wchar_t is?

Question

The following C++ program compiles without errors:

void f(char){}
void f(signed char){}
void f(unsigned char){}
int main(){}  

The wchar_t version of the same program does not:

void f(wchar_t){}
void f(signed wchar_t){}
void f(unsigned wchar_t){}
int main(){}

error: redefinition of ‘void f(wchar_t)’
void f(signed wchar_t){}

It seems that wchar_t is unsigned.
Why is there an inconsistency in overloading?

Answer 1

The chars are all distinct types and can be overloaded

[basic.fundamental] / 1

[...] Plain char, signed char, and unsigned char are three distinct types, collectively called narrow character types. [...]

wchar_t is also a distinct type, but it cannot be qualified with signed or unsigned, which can only be used with the standard integer types.

[dcl.type] / 2

As a general rule, at most one type-specifier is allowed in the complete decl-specifier-seq of a declaration or in a type-specifier-seq or trailing-type-specifier-seq. The only exceptions to this rule are the following:

[...]

signed or unsigned can be combined with char, long, short, or int.

[dcl.type.simple] / 2

[...] Table 9 summarizes the valid combinations of simple-type-specifiers and the types they specify.

The signedness of wchar_t is implementation defined:

[basic.fundamental] / 5

[...] Type wchar_t shall have the same size, signedness, and alignment requirements (3.11) as one of the other integral types, called its underlying type.