Should useless type qualifiers on return types be used, for clarity?

Question

Our static analysis tool complains about a \"useless type qualifier on return type\" when we have prototypes in header files such as:

const int foo();
         


        

Answer 1

Yes. I would advise writing code "explicitly", because it makes it clear to anyone (including yourself) when reading the code what you meant. You are writing code for other programmers to read, not to please the whims of the compiler and static analysis tools!

(However, you do have to be careful that any such "unnecessary code" does not cause different code to be generated!)

Some examples of explicit coding improving readability/maintainability:

• I place brackets around portions of arithmetic expressions to explicitly specify what I want to happen. This makes it clear to any reader what I meant, and saves me having to worry about (or make ay mistakes with) precedence rules:

  int a = b + c * d / e + f;      // Hard to read- need to know precedence
  int a = b + ((c * d) / e) + f;  // Easy to read- clear explicit calculations
  

• In C++, if you override a virtual function, then in the derived class you can declare it without mentioning "virtual" at all. Anyone reading the code can't tell that it's a virtual function, which can be disastrously misleading! However you can safely use the virtual keyword:

  virtual int MyFunc()

  and this makes it clear to anyone reading your class header that this method is virtual. (This "C++ syntax bug" is fixed in C# by requiring the use of the "override" keyword in this case - more proof if anyone needed it that missing out the "unnecessary virtual" is a really bad idea)

These are both clear examples where adding "unnecessary" code will make the code more readable and less prone to bugs.