When returning a boolean, why use "!!"

Question

I saw some C++ code like this:

bool MyProject::boo(...)
{
  bool fBar = FALSE;
  ....
  return !!fBar;
}

I couldn't think of any difference between returning fBar directly in this case and returning !!fBar. How two negatives make a difference?

Thank you

Answer 1

In your example, there's no difference between returning fBar and returning !!fBar.

In other cases, e.g. when a user-defined type such as BOOL (typedef-ed to be int) is used, the !! construct has the effect of coercing any non-zero value to true; i.e. !!fBar is equivalent to fBar ? true : false. This can make a difference if fBar can be 5 and you want to compare it against TRUE, which is defined to be (BOOL)1.

Answer 2

This is typically done to avoid compiler warnings in situations when a non-bool value has to be converted to bool type. Some compilers (like MSVC++) issue a "performance" warning when a non-bool value is implicitly converted to bool. One way to suppress this warning is to use an explicit conversion. Another way is to use the !! combination.

However, in your case the argument of return is already declared as a bool, meaning that the above reasoning does not apply. (Are you sure it was bool and not, say, BOOL?). In that case there's no meaningful explanation for that !!.