Is initializer evaluated after memory allocation in new expression?

Question

Consider the code

auto p = new T( U(std::move(v)) );

The initializer is then U(std::move(v)). Let's assume that T( U(std::move(v)) ) does not throw. If the initializer is evaluated after the underlying memory allocation, the code is then strong-exception-safe. Otherwise, it is not. Had memory allocation thrown, v would have already been moved. I'm therefore interested in the relative order between memory allocation and initializer evaluation. Is it defined, unspecified, or what?

Answer 1

Yes, the initialisation is evaluated after the allocation. Quoting C++17 (N4659) [expr.new] 8.3.4/19:

The invocation of the allocation function is sequenced before the evaluations of expressions in the new-initializer. Initialization of the allocated object is sequenced before the value computation of the new-expression.