Misra violation 12.6

Question

How to get rid of MISRA violation on following statement

typedef unsigned char boolean;

boolean A, B;

A = !B;

Operand of logical ! operator is not an 'effectively Boolean' expression. MISRA-C:2004 Rule 12.6; REFERENCE - ISO:C90-6.3.3.3 Unary Arithmetic Operators

Answer 1

If you read rule 12.6 it says "check Boolean Expressions" in the appendix. There we can read

"Boolean-by-enforcement values can be introduced by implementing a specific type enforcement mechanism using a tool. A Boolean type could be associated with a specific typedef, and would then be used for any objects that are Boolean. This could bring many benefts, especially if the checking tool can support it, and in particular it could help avoid confusion between logical operations and integer operations."

MISRA-C:2004 assumes C90, and in C90 there is no bool type, you have to typedef it yourself, like you have done. Since your intention is to have a type which is effectively boolean, the code is just fine. In fact, your code follows MISRA recommendations beyond the mandatory ones.

The problem lies with your tool: it either does not support to allow a specific boolean type as per MISRA recommendations, or it is misconfigured.

Answer 2

Simple... don't use ! on things that aren't booleans. Just because your typedef is named boolean doesn't mean it is; it's still an unsigned char.

You could write:

if (b == 0) A = 1;
else A = 0;

I don't think MISRA allows ternary operators (could be wrong; not an expert) but if it does, you could write:

A = (b == 0) ? 1 : 0;