How it works...

The new constexpr-if works exactly like usual if-else constructs. The difference is that the condition that it tests has to be evaluated at compile time. All runtime code that the compiler creates from our program will not contain any branch instructions from constexpr-if conditionals. One could also put it that it works in a similar manner to preprocessor #if and #else text substitution macros, but for those, the code would not even have to be syntactically well-formed. All the branches of a constexpr-if construct need to be syntactically well-formed, but the branches that are not taken do not need to be semantically valid.

In order to distinguish whether the code should add the value x to a vector or not, we use the type trait std::is_same. An expression std::is_same<A, B>::value evaluates to the Boolean value true if A and B are of the same type. The condition used in our recipe is std::is_same<T, std::vector<U>>::value, which evaluates to true if the user specialized the class on T = std::vector<X> and tries to call add with a parameter of type U = X.

There can, of course, be multiple conditions in one constexpr-if-else block (note that a and b have to depend on template parameters and not only on compile-time constants):

if constexpr (a) {
 // do something
} else if constexpr (b) {
 // do something else 
} else {
 // do something completely different
}

With C++17, a lot of meta programming situations are much easier to express and to read.