问题
Can C++11 compilers (and do they) notice that a function is a constexpr and treat them as such even if they are not declared to be constexpr?
I was demonstrating the use of constexpr to someone using the example straight from the Wikipedia:
int get_five() {return 5;}
int some_value[get_five() + 7]; // Create an array of 12 integers. Ill-formed C++
To my surprise the compiler was OK with it. So, I further changed get_five( ) to take a few int parameters, multiply them and return the result while still not being explicitly declared to be constexpr. The compiler was OK with that as well. It seems that if the compiler can do this there isn't much point to having the restrictions that are required in order to explicitly declare something constexpr.
回答1:
On a properly-functioning C++11 compiler, your code would be rejected.
Based on its being accepted, you're almost certainly using gcc (or something that closely emulates its bugs). gcc [depending somewhat on flags] can accept array sizes that aren't constant by any measure (e.g., depend on run-time input from the user) because they support an analog of C99 variable-length arrays in C++.
回答2:
A compiler can detect whether or not a function could have been declared with constexpr even when they haven't, for optimization purposes (i.e. computing the result of a function at compile-time). Compilers did that prior to C++11.
But for use in places that requires constant expressions, such as template parameters of integral type, it is against the standard to allow calls to functions that are not declared with the constexpr keyword.
来源：https://stackoverflow.com/questions/15749825/implicit-constexpr