Quick Q: How is “=default” different from “{}” for default constructor and destructor?

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Quick A: The "= default" keeps the type trivial.

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How is “=default” different from “{}” for default constructor and destructor?

I originally posted this as a question only about destructors, but now I'm adding consideration of the default constructor. Here's the original question:

If I want to give my class a destructor that is virtual, but is otherwise the same as what the compiler would generate, I can use =default:

class Widget {
public:
   virtual ~Widget() = default;
};

But it seems that I can get the same effect with less typing using an empty definition:

class Widget {
public:
   virtual ~Widget() {}
};

Is there any way in which these two definitions behave differently?

Based on the replies posted for this question, the situation for the default constructor seems similar. Given that there is almost no difference in meaning between "=default" and "{}" for destructors, is there similarly almost no difference in meaning between these options for default constructors? That is, assuming I want to create a type where the objects of that type will be both created and destroyed, why would I want to say

Widget() = default;

instead of

Widget() {}

?

I apologize if extending this question after its original posting is violating some SO rules. Posting an almost-identical question for default constructors struck me as the less desirable option.

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Dave Johansen said on Apr 13, 2015 11:19 AM:

The actual SO article can be found at:
http://stackoverflow.com/questions/13576055/how-is-default-different-from-for-default-constructor-and-destructor