CODEWITHSUNDEEP
c++stlg++clang++
assassin5615
assassin5615

Reputation: 87

why is_standard_layout gives true here?

my understanding is that not having any virtual method is one of requirements for standard layout. however, following code prints out true while there is a virtual method in struct Cancel.

if I remove the virtual method. is_stand_layout returns false which makes sense, as Cancel has members with different access control.

I tested this on gcc 10.2, 9.2 9.3 and clang 11.0, the results are the same.

is this a bug? or is there something I am missing about standard layout?

#include <variant>
#include <iostream>

struct NewOrder
{
    int i;
};
struct Cancel
{
    virtual ~Cancel() {}
    int j;
private:
    int i;
}; 
int main() {
  std::cout << std::is_standard_layout_v<std::variant<NewOrder, Cancel>> << std::endl;
  return 0;
}

Upvotes: 4

Views: 257

Answers (1)

eerorika
eerorika

Reputation: 238401

is this a bug?

No.

is there something I am missing

You're outputting std::is_standard_layout_v<std::variant<NewOrder, Cancel>> which determines whether std::variant<NewOrder, Cancel> is standard layout rather than whether Cancel is standard layout.

why is_standard_layout gives true here?

Whether std::variant<NewOrder, Cancel> is standard layout or not doesn't necessarily depend on whether the template type arguments are themselves standard layout. The standard doesn't specify whether std::variant is or isn't standard layout. The output that you see is a side-effect of how std::variant was implemented in the standard library that you used.

Upvotes: 2

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