Concerning this question
Convert from uint8_t * to char * in C
The suggestion was simply to cast with (char*). Why is this acceptable in this context? Or did the responder just take the question literally?
Casting unsigned (depending on the implementation) to signed feels dubious.
In C, it is allowed to access a variable of a given type through a pointer to the signed or unsigned version of that that type.
This is specified in section 6.5p7 of the C standard:
An object shall have its stored value accessed only by an lvalue expression that has one of the following types:
- a type compatible with the effective type of the object,
- a qualified version of a type compatible with the effective type of the object,
- a type that is the signed or unsigned type corresponding to the effective type of the object,
- a type that is the signed or unsigned type corresponding to a qualified version of the effective type of the object,
- an aggregate or union type that includes one of the aforementioned types among its members (including, recursively, a member of a subaggregate or contained union), or
- a character type.
A uint8_t is an alias for an unsigned char (at least, on systems where CHAR_BIT is 8), so it is allowed to convert an uint8_t * to a char * and subsequently dereference the resulting pointer.
Note also that, per the last bullet point above, using a pointer any character type is allowed to access any kind of object.