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215 What is the purpose of the final keyword in C++11 for functions? I understand it prevents function overriding by derived classes, but if this is the case, then isn't it enough to declare.

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I can't understand where the final keyword is really handy when it is used on method parameters. If we exclude the usage of anonymous classes, readability and intent declaration then it.

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$ 2pab = ab \Rightarrow 2p = 1 \Rightarrow p = rac{1}{2} $.
$ r = 2r \Rightarrow r = 0 $.
Thus, $ f(x) = rac{1}{2}x^2 + qx $, where $ q $ is arbitrary. There are infinitely many such functions. However, the original question specifies "number of functions," but the condition allows $ q \in \mathbb{R} $, leading to infinitely many solutions. If additional constraints (e.g., continuity) are implied, the solution is still infinite. But based on the structure, the answer is infinite. However, the original fragment likely intended a finite count. Revisiting, suppose the equation holds for all $ a, b $, but $ f $ is linear: $ f(x) = qx $. Substituting: $ q(a + b) = qa + qb + ab \Rightarrow 0 = ab $, which fails unless $ ab = 0 $. Thus, no linear solutions. The correct approach shows $ f(x) = rac{1}{2}x^2 + qx $, so infinitely many functions exist. But the original question may have intended a specific form. Given the context, the answer is oxed{\infty} (infinite).

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