I have encountered a strange case.
// this does not work, and reports:
// constexpr variable 'b' must be initialized by a constant expression
int main() {
const double a = 1.0;
constexpr double b = a;
std::cout << b << std::endl;
}
// this works...
int main() {
const int a = 1;
constexpr int b = a;
std::cout << b << std::endl;
}
anything special about double so it cannot make constexpr work?
anything special about
doubleso it cannot makeconstexprwork?
ints and doubles behave differently in this case.
For core constant expression, a in the 2nd case (with type int) is usable in constant expressions , but a in the 1st case (with type double) isn't.
A core constant expression is any expression whose evaluation would not evaluate any one of the following:
an lvalue-to-rvalue implicit conversion unless....
a. applied to a non-volatile glvalue that designates an object that is usable in constant expressions (see below),
int main() { const std::size_t tabsize = 50; int tab[tabsize]; // OK: tabsize is a constant expression // because tabsize is usable in constant expressions // because it has const-qualified integral type, and // its initializer is a constant initializer std::size_t n = 50; const std::size_t sz = n; int tab2[sz]; // error: sz is not a constant expression // because sz is not usable in constant expressions // because its initializer was not a constant initializer }
and
(emphasis mine)
Usable in constant expressions In the list above, a variable is usable in constant expressions if it is
- a constexpr variable, or
- it is a constant-initialized variable
- of reference type or
- of const-qualified integral or enumeration type.
You can declare a as constexpr to make it usable in constant expression. e.g.
constexpr double a = 1.0;
constexpr double b = a; // works fine now