Why do `char`, `signed char`, and `unsigned char` behave differently from `int` and its variants in C template specializations?

char!=(signed char), char!=(unsigned char) [duplicate]

In certain scenarios, C treats char, signed char, and unsigned char as distinct types, unlike int and its variants. This behavior stems from the nature of character types in C , as outlined by the C standard:

Plain char, signed char, and unsigned char are three distinct types.

While int is typically synonymous with signed int, char does not have the same relationship with signed char or unsigned char. This means that template specializations for these types will be instantiated separately.

In the provided code, three template specializations are created:

• isX::ikIsX (for signed char)
• isX::ikIsX (for unsigned char)
• isX::ikIsX (for plain char)

However, for int and its variants, only one template specialization is created because they are all considered to be the same type unless explicitly specified as signed or unsigned.

Therefore, in the given example, the output will show three different values for the isX::ikIsX type aliases for int8, uint8, and char, while it will show only one value for int32, uint32, and int.

This distinction is crucial to understand when working with template metaprogramming and type introspection in C .

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