With a constexpr-specified function foo_constexpr I have code such as shown below:
const auto x = foo_constexpr(y); static_assert(x==0); Under which circumstances could the code then fail to compile, when the declaration of x is changed to constexpr? (After all, x must already be a constant expression for use in the static_assert.) That is:
constexpr auto x = foo_constexpr(y); static_assert(x==0); In general, it can fail to compile when the execution of foo_constexpr violates a requirement of constant expressions. Remember, a constexpr function is not a function that is always a constant expression. But rather it is a function that can produce a constant expression for at lease one input! That's it.
So if we were to write this perfectly legal function:
constexpr int foo_constexpr(int y) { return y < 10 ? 2*y : std::rand(); } Then we'll get:
constexpr int y = 10; const auto x1 = foo_constexpr(y); // valid, execution time constant constexpr auto x2 = foo_constexpr(y); // invalid, calls std::rand But of course, if x is already usable in a constant expression (such as a static assertion), changing to constexpr cannot cause a failure to occur.
constexpr. Alas I've not yet been able to reduce the code to something useful to share.
static_assert and just switching const to constexpr causes failure, then there's potential from something fishy here.
y? Even if the function isconstexprit can still be called at runtime with a runtime value.foo_constexprcan still be part of a constant expression in that case: as long as it doesn't use the value ofywhen calculating the result.yis and how it's used.