bool operator idiom

http://en.wikibooks.org/wiki/More_C++_Idioms/Safe_bool

흐 - 음. 

선 기록 후 정리

struct Testable
{
    operator bool() const {
          return false;
    }
};
struct AnotherTestable
{
    operator bool() const {
          return true;
    }
};
int main (void)
{
  Testable a;
  AnotherTestable b;
  if (a == b) { /* blah blah blah*/ }
  if (a < 0) { /* blah blah blah*/ }
  // The above comparisons are accidental and are not intended but the compiler happily compiles them.
  return 0;
}

Solution

class Testable {
    bool ok_;
    typedef void (Testable::*bool_type)() const;
    void this_type_does_not_support_comparisons() const {}
  public:
    explicit Testable(bool b=true):ok_(b) {}
 
    operator bool_type() const {
      return ok_ ? 
        &Testable::this_type_does_not_support_comparisons : 0;
    }
};
template <typename T>
bool operator!=(const Testable& lhs, const T&) {
    lhs.this_type_does_not_support_comparisons();	
    return false;
}
template <typename T>
bool operator==(const Testable& lhs, const T&) {
    lhs.this_type_does_not_support_comparisons();
    return false;
}
class AnotherTestable ... // Identical to Testable.
{};
int main (void)
{
  Testable t1;
  AnotherTestable t2;
  if (t1) {} // Works as expected
  if (t2 == t1) {} // Fails to compile
  if (t1 < 0) {} // Fails to compile
 
  return 0;
}

In C++11, the explicit keyword can now be applied to conversion operators. As with constructors, it prevents the use of those conversion functions in implicit conversions. However, language contexts that specifically require a boolean value (the conditions of if-statements and loops, as well as operands to the logical operators) count as explicit conversions and can thus use a bool conversion operator.

struct Testable
{
    explicit operator bool() const {
          return false;
    }
};
 
int main()
{
  Testable a, b;
  if (a)      { /*do something*/ }  // this is correct
  if (a == b) { /*do something*/ }  // compiler error
}