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We need a few clarifications about the computation of underlying types for constants according to MISRA-C++-2008.
Please note that the following questions, even though using concrete examples for clarity, are meant to be general; in particular, here we are not asking how the examples should be changed in order to make them compliant with MISRA C++.
Question 1:
Given the following context:
Code: enum Colours { RED, BLUE, GREEN };
static const Colours red = RED;
In section 6.5.0 of MISRA C++ (page 59) it is said:
Quote: Additive operators
[...]
The result is a cvalue expression whose underlying type is as defined by the underlying type conversions.
The underlying type conversions (page 57) state that if one of the arguments has enum type then the expression has enum type. Therefore, both expressions should be cvalues having underlying type enum Colours.
However, both expressions are integer constant expressions according to the C++ standard and hence the MISRA C++ rule about the underlying type of constants (page 60) seems to be relevant too:
Quote:Constant expressions
The result is not a cvalue expression. The underlying type for a constant expression “e†with a value “v†will have the same signedness as “eâ€, and a magnitude given by the underlying type of a single integer-literal with the same value as “vâ€.
According to this rule, we should consider integer literals with the actual values of the expressions (i.e., 1), thereby obtaining non-cvalue expressions having underlying type signed char.
To summarize:
1a) we would like to know if the MISRA rule about constant expressions should be applied only to expressions whose MISRA underlying type (computed as if they were not constants) is "effectively" integral, i.e., different from bool, plain char and enum;
1b) if that is the case (i.e., if red + 1 has underlying type enum), then what about its cvalue-ness? Is red + 1 a cvalue or not?
Question 2:
What is the underlying type of expression ?
In section 6.5.0 of MISRA C++ (page 58) it is said:
Quote:Literals
The underlying type of an integral literal is the smallest fundamental type of the appropriate sign required to store its value. For example, the underlying type of the literal 128 is S16. The result is not a cvalue.
So, the underlying type of 1L should be signed char.
Is this really meant to be different from the underlying type computed according to MISRA-C-2004? For instance, in the examples for rule 10.1 (page 43 of MISRA-C-2004) we have the following line:
Code: u8a = 5UL; /* not compliant */
To summarize: we would like to know if the MISRA rule about constant expressions should be applied only to expressions whose MISRA underlying type (computed as if they were not constants) is one of signed char, unsigned char, short, unsigned short, int, unsigned int, thereby disregarding the "big" integer types.
Question 3:
Assuming that the answer to the Question 2 above is that the underlying type of 1UL is unsigned long, what is the cvalue-ness of ?
If 1UL has underlying type unsigned long, then the rule about constants has not been applied. Hence, the only relevant rule is the one about the unary operators. In section 6.5.0 of MISRA C++ (page 58) it is said:
Quote:Unary expressions
[...]
- cast-expression
The result is a cvalue expression whose underlying type is that of the cast-expression.
Therefore, -1UL is not a cvalue. Is this meant, or is it the case that any ICE expression (no matter if bool, char, enum or effectively integer) is not a cvalue?
Thanks in advance for any clarification.
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