Reputation: 60341
What does the C++11 iso standard say about such an expression :
class MyClass
{
public:
constexpr int test()
{
return _x;
}
protected:
int _x;
};
_x
is a non-const used in a constexpr
: will it produce an error, or will the constexpr
be simply ignored (as when we pass a non-const parameter) ?
Upvotes: 0
Views: 129
Reputation: 476930
It's perfectly fine, though somewhat useless:
constexpr int n = MyClass().test();
Since MyClass
is an aggregate, value-initializing it like that will value-initialize all members, so this is just zero. But with some polish this can be made truly useful:
class MyClass
{
public:
constexpr MyClass() : _x(5) { }
constexpr int test() { return _x; }
// ...
};
constexpr int n = MyClass().test(); // 5
Upvotes: 5
Reputation: 227370
If the expression does not resolve to a constant expression, then it cannot be used as such. But it can still be used:
#include <array>
constexpr int add(int a, int b)
{
return a+b;
}
int main()
{
std::array<int, add(5,6)> a1; // OK
int i=1,
int j=10;
int k = add(i,j); // OK
std::array<int, add(i,j)> a2; // Error!
}
Upvotes: 1