On 10/05/06, Joe Buck <[EMAIL PROTECTED]> wrote:
But then I just thought of another case:
template <typename Container>
struct foo {
foo(const Container& bar) { if (bar.size() >= 0) use(bar); }
};
For any STL-compliant container the test is redundant. But if
we put in a rule saying to suppress the warning if the type
depends on a template, we lose the warning in this case as well;
after all, nothing stops someone from writing
class C {
public:
int size() const;
...
};
void use(const C&);
My STL may be a bit rusty...if you don't define size(), what is its
return type?. Also, there should be no warning in this example, am I
wrong?
