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When C++14 lifted restrictions on constexpr it seemed to include the following (copied from Wikipedia):
Expressions may change the value of an object if the lifetime of that object began within the constant expression function. This includes calls to any non-const constexpr-declared non-static member functions.
That seems to imply that you could create an object using new and as long as you delete it within the expression, then it would be allowed.
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constexpr indicates that the value, or return value, is constant and, where possible, is computed at compile time. A constexpr integral value can be used wherever a const integer is required, such as in template arguments and array declarations.
A const int var can be dynamically set to a value at runtime and once it is set to that value, it can no longer be changed. A constexpr int var cannot be dynamically set at runtime, but rather, at compile time. And once it is set to that value, it can no longer be changed.
A call to a constexpr function produces the same result as a call to an equivalent non- constexpr function , except that a call to a constexpr function can appear in a constant expression. The main function cannot be declared with the constexpr specifier.
Short answer: static_assert(false) should never appear in a constexpr if expression, regardless of whether it's in a template function or whether it's in the discarded branch.
Language lawyer answer. All references to N3797.
7.1.5/5 states:
For a non-template, non-defaulted
constexprfunction or a non-template, non-defaulted, non-inheritingconstexprconstructor, if no argument values exist such that an invocation of the function or constructor could be an evaluated subexpression of a core constant expression (5.19), the program is ill-formed; no diagnostic required.
Jumping over to 5.19, we see:
A conditional-expression
eis a core constant expression unless the evaluation ofe, following the rules of the abstract machine (1.9), would evaluate one of the following expressions:
... [lots of bullets]...
a new-expression (5.3.4);
a delete-expression (5.3.5);
... [lots more bullets]...
So no: a program containing a constexpr function with an unconditional invocation of new or delete in it is ill-formed, no diagnostic required. (I'd be surprised, however, if any half-decent compiler failed to diagnose such invocations of new or delete in a constexpr function, required or not.)
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