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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?
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