CODEWITHSUNDEEP
c++c++11language-lawyerconstexprc++14
Khurshid
Khurshid

Reputation: 2724

can have definition variable of non-literal type in constexpr function body c++14?

I think that in C++14 more restrictives are removed from constexpr. But according N3797 7.1.5 3-punct:

The definition of a contexpr function shall satisfy the following constraints:

I know why static, thread storage duration variables are disallowed, but I don't see any reason, why ONLY allowed definition of a variable of literal type ?

Or I'm not understand the standard.

I'm not sure, but according standard following errors should be created even C++14:

struct point{
constexpr point(): x(0), y(0){}
constexpr point(int x_, int y_): x(x_),y(y_){}
constexpr int hypot()const { return x*x + y*y; }
int x,y;
};

constexpr int hypot(int x, int y) {  
   point p{x,y}; //error, because p - is not literal type.
   return p.hypot();
}

// error, because return type is not literal.
constexpr point  getPoint(int x, int y) { return {x,y}; }

// error, because parameter is not literal.
constexpr int hypot(point p) { return p.hypot(); }

Q: If really above errors would happen, why these restrictions are not removed?

Upvotes: 17

Views: 46787

Answers (1)

Casey
Casey

Reputation: 42584

literal type is defined in 3.9/10:

A type is a literal type if it is:

  • void; or

  • a scalar type; or

  • a reference type; or

  • an array of literal type; or

  • a class type (Clause 9) that has all of the following properties:

    • it has a trivial destructor,

    • it is an aggregate type (8.5.1) or has at least one constexpr constructor or constructor template that is not a copy or move constructor, and

    • all of its non-static data members and base classes are of non-volatile literal types

So your struct point is a literal type and your sample code is valid C++1y.

As to why constexpr functions are restricted to variables of literal type, they are the only types that are guaranteed to be interpretable at compile time.

Upvotes: 28

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