Monday, March 29, 2010

const overloading -- needed in these cases

volatile and const are 2 keywords treated specially.
---
typedef .... my_iterator;
typedef .... my_const_interator;

my_iterator begin();
my_const_iterator begin() const;

Are we looking at method-overloading? yes due to the const. Note the return type is ignored when overloading.

The c++ const correctness article explains the motivation of const overloading. Here's my explanation. Suppose you have 2 variables of MyType, one const var (call it varC), the other non-const var (call it var2). Suppose MyType has a non-const getter method. How would you call getter on varC ? Won't compile[3]. You need to "const overload" the getter. So the 2 overloaded getters differ only in constness. How does compiler resolve the call? varC.getIt() would call the const getter; var2.getIt() would call the mutable getter.

[3] varC is a const "handle" on the object, so compiler promises not to modify the object state through this handle. Compiler won't pass the unsafe (ie mutable) message getIt() to the object through this handle. Doing so creates the possibility of state mutation via a const handle.

http://www.parashift.com/c++-faq-lite/const-correctness.html#faq-18.12
gave a good coding convention, and gave detailed and simple examples.

Fred const& operator[] (unsigned index) const; ← subscript operators often come in pairs
Fred& operator[] (unsigned index); ← subscript operators often come in pairs

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