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I think that your question is actually two questions: <ul> <li>What are the implications of returning a const.</li> <li>What are the implications of returning a reference.</li> </ul> To give you a better answer, I will explain a little more about both concepts. Regarding the const keyword The const keyword means that the object cannot be modified through that variable, for instance: <pre><code>MyObject *o1 = new MyObject; const MyObject *o2 = o1; o1->set(...); // Will work and will change the instance variables. o2->set(...); // Won't compile. </code></pre> Now, the const keyword can be used in three different contexts: <ul> <li>Assuring the caller of a method that you won't modify the object</li> </ul> For example: <pre><code>void func(const MyObject &o); void func(const MyObject *o); </code></pre> In both cases, any modification made to the object will remain outside the function scope, that's why using the keyword const I assure the caller that I won't be modifying it's instance variables. <ul> <li>Assuring the compiler that a specific method do not mutate the object</li> </ul> If you have a class and some methods that "gets" or "obtains" information from the instance variables without modifying them, then I should be able to use them even if the const keyword is used. For example: <pre><code>class MyObject { ... public: void setValue(int); int getValue() const; // The const at the end is the key }; void funct(const MyObject &o) { int val = o.getValue(); // Will compile. a.setValue(val); // Won't compile. } </code></pre> <ul> <li>Finally, (your case) returning a const value</li> </ul> This means that the returned object cannot be modified or mutated directly. For example: <pre><code>const MyObject func(); void func2() { int val = func()->getValue(); // Will compile. func()->setValue(val); // Won't compile. MyObject o1 = func(); // Won't compile. MyObject o2 = const_cast<MyObject>(func()); // Will compile. } </code></pre> More information about the const keyword: <a href="http://www.parashift.com/c++-faq-lite/const-correctness.html" rel="nofollow noreferrer">C++ Faq Lite - Const Correctness</a> Regarding references Returning or receiving a reference means that the object will not be duplicated. This means that any change made to the value itself will be reflected outside the function scope. For example: <pre><code>void swap(int &x, int &y) { int z = x; x = y; y = z; } int a = 2; b = 3; swap(a, b); // a IS THE SAME AS x inside the swap function </code></pre> So, returning a reference value means that the value can be changed, for instance: <pre><code>class Foo { public: ... int &val() { return m_val; } private: int m_val; }; Foo f; f.val() = 4; // Will change m_val. </code></pre> More information about references: <a href="http://www.parashift.com/c++-faq-lite/value-vs-ref-semantics.html" rel="nofollow noreferrer">C++ Faq Lite - Reference and value semantics</a> Now, answering your questions <pre><code>const int exampleOne(); </code></pre> Means the object returned cannot change through the variable. It's more useful when returning objects. <pre><code>int& exampleTwo(); </code></pre> Means the object returned is the same as the one inside the function and any change made to that object will be reflected inside the function. <pre><code>const int& exampleThree(); </code></pre> Means the object returned is the same as the one inside the function and cannot be modified through that variable.
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1. POWhich are the implications of return a value as constant, reference and constant reference in C++?
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1. COGood overview, but you neglected to mention the most important difference between reference and value return types: returning a reference to a local or temporary object will compile but cause crashes because the value is destroyed before it gets to the calling code. Also, I think you want "func().getValue()" and "func().setValue()" in your 4th code snippet.
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