Robert, I see Andrey's proposal aimed to replace the enable_if which is based on SFINAE and greatly simplify the metaprogramming machinery. The code you present if fine and it might be optimized away by the compiler, but it might produce compilation errors, since all runtime branches of if-statement must be compilable without errors. SFINAE aimes to work around it, like we can enable some special treatment if the provided code "would compile" without errors.

On the other hand using static if we might inspect the exposed type system of some type T. Let's say we would like to unify some different types using a traits class. Out traits class should expose value_type of the inspected type (say we have a boost::shared_ptr and std::vector as input). boost::shared_ptr contains a typedef of underlying type which is named element_type and std::vector names the underlying type value_type. Out traits type should homogenize these two types and provide a value_type member, which contains the underlying type of either shared_ptr or std::vector. It can be easily "calculated/specialized/inspected" with static if construct. How are you going to solve this problem with the runtime if without using enable_if , overloads and template specializations? 

IMO static if is aimed to simplify lot's of code, make it readable and understandable, this is what it is about.

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Hope that helps,

Ovanes