How to implement a member function once for both non-const and const overloads? [duplicate]

Question

Sometimes I find myself adding overloads that have the same implementation, with const in qualifier and return value being the only difference:

struct B {}; struct A { const B& get(int key) const { if (auto i = map.find(key); i != map.end()) return i->second; throw std::runtime_error{""}; } B& get(int key) { if (auto i = map.find(key); i != map.end()) return i->second; throw std::runtime_error{""}; } private: std::unordered_map<int, B> map; };

Is there an idiomatic way to write the implementation only once and get rid of copy-paste that is better than const_cast:

const B& get(int key) const { return const_cast<A*>(this)->get(key); }

?

Good question. Definitely a duplicate. An adequate pattern for this is yet to emerge. Scott Meyers uses the const_cast technique. — Bathsheba
– Bathsheba, Commented Sep 21, 2018 at 9:47

P.W · Accepted Answer · 2018-09-21 10:00:23Z

1

Scott Meyers' advice:

When const and non-const member functions have essentially identical implementations, code duplication can be avoided by having the non-const version call the const version.

answered Sep 21, 2018 at 10:00

P.W

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4 Comments

bipll Over a year ago

Not always feasible:

struct Counterexample { int meow() { return 42; } int meow() const { return 314; } int nyan() { return meow(); } int nyan() const { return meow(); } };

P.W Over a year ago

Agree. But it may be useful in the OP context.

Dev Null Over a year ago

so it's essentially (almost) the same as my example with const_cast

P.W Over a year ago

Two casts are used. One is a static_cast to add const to *this’s type, so that the const version will be called. and the other is const_cast to cast away const on the return type. And these casts are used in the non-const version.

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