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C++ static_cast runtime overhead

开发者 https://www.devze.com 2023-03-14 14:59 出处:网络
See the code below. a) Does, in this case (simple inheritance, no virtual members), the static cast in B::df() have any overhead (whatsoever)? I found some conflicting answers to similar questions, t

See the code below.

a) Does, in this case (simple inheritance, no virtual members), the static cast in B::df() have any overhead (whatsoever)? I found some conflicting answers to similar questions, that's why I am asking...

b) I was thinking about making const M1 * func private in A and introducing a new private field const M2 * func into B to avoid the cast, but it kind of complicates things up and makes use of smart pointers more difficult. Do you see a better way to avoid the cast?


class M1 {
public:
    double f()开发者_如何转开发 const;
};

class M2 : public M1 { public: double df() const; };

class A { protected: const M1 * func; public: A(const M1 * p); ~A(); double f() const; };

class B : public A { public: B(const M2 * p); double df() const; };


double M1::f() const { return 1973.0; }
double M2::df() const { return 0.0; }

A::~A() { delete func; } A::A(const M1 * p) : func(p) {} double A::f() const { return func->f(); }

B::B(const M2 * p) : A(p) {} double B::df() const { return static_cast<const M2*>(func)->df(); }


static_cast<T>(e) is equivalent to creating an invented temporary variable v in the following way:

T v(e); //where T is an arbitrary type  and e is an arbitrary expression.

The runtime cost of a static_cast is exactly the cost of the above statement

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