Is unsigned char always promoted to int?
c, c99, integer-promotion, type-conversion, type-promotion
Solution
are implementations allowed to promote them to unsigned int?
Implementations will promote to `unsigned int` if not all `unsigned char` values are representable in an `int` (as ruled by 6.2.5p9 in C99). See below for implementation examples.
If so, does that imply that an implementation could theoretically have an unsigned char value which is not in the subrange of an int?
Yes, example: DSP cpu with `CHAR_BIT` 16 or 32.
For example, TI C compiler for TMS320C55x: `CHAR_BIT` is 16 and `UCHAR_MAX` 65535, `UINT_MAX` 65535 but `INT_MAX` 32767.
http://focus.ti.com/lit/ug/spru281f/spru281f.pdf
Problem
Suppose the following: ``` unsigned char foo = 3; unsigned char bar = 5; unsigned int shmoo = foo + bar; ``` Are `foo` and `bar` values guaranteed to be promoted to `int` values for the evaluation of the expression `foo + bar` -- or are implementations allowed to promote them to `unsigned int`? In section 6.2.5 paragraph 8: For any two integer types with the same signedness and different integer conversion rank (see 6.3.1.1), the range of values of the type with smaller integer conversion rank is a subrange of the values of the other type. In section 6.2.5 paragraph 9: If an `int` can represent all values of the original type, the value is converted to an `int`; otherwise, it is converted to an `unsigned int`. The guarantee that an integer type with smaller integer conversion rank has a range of values that is a subrange of the values of the other type seems dependent on the signedness of the integer type. - `signed char` corresponds to `signed int` - `unsigned char` corresponds to `unsigned int` Does this mean that the value of an `unsigned char` is only guaranteed to be in the subrange of `unsigned int` and not necessarily `int`? If so, does that imply that an implementation could theoretically have an `unsigned char` value which is not in the subrange of an `int`?