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

This is a completely different question when asking about constructors than destructors.

If your destructor is virtual, then the difference is negligible, as Howard pointed out. However, if your destructor was non-virtual, it's a completely different story. The same is true of constructors.

Using = default syntax for special member functions (default constructor, copy/move constructors/assignment, destructors etc) means something very different from simply doing {}. With the latter, the function becomes "user-provided". And that changes everything.

This is a trivial class by C++11's definition:

struct Trivial
{
  int foo;
};

If you attempt to default construct one, the compiler will generate a default constructor automatically. Same goes for copy/movement and destructing. Because the user did not provide any of these member functions, the C++11 specification considers this a "trivial" class. It therefore legal to do this, like memcpy their contents around to initialize them and so forth.

This:

struct NotTrivial
{
  int foo;

  NotTrivial() {}
};

As the name suggests, this is no longer trivial. It has a default constructor that is user-provided. It doesn't matter if it's empty; as far as the rules of C++11 are concerned, this cannot be a trivial type.

This:

struct Trivial2
{
  int foo;

  Trivial2() = default;
};

Again as the name suggests, this is a trivial type. Why? Because you told the compiler to automatically generate the default constructor. The constructor is therefore not "user-provided." And therefore, the type counts as trivial, since it doesn't have a user-provided default constructor.

The = default syntax is mainly there for doing things like copy constructors/assignment, when you add member functions that prevent the creation of such functions. But it also triggers special behavior from the compiler, so it's useful in default constructors/destructors too.

The important difference between

class B {
    public:
    B(){}
    int i;
    int j;
};

and

class B {
    public:
    B() = default;
    int i;
    int j;
};

is that default constructor defined with B() = default; is considered not-user defined. This means that in case of value-initialization as in

B* pb = new B();  // use of () triggers value-initialization

special kind of initialization that doesn't use a constructor at all will take place and for built-in types this will result in zero-initialization. In case of B(){} this won't take place. The C++ Standard n3337 § 8.5/7 says

To value-initialize an object of type T means:

— if T is a (possibly cv-qualified) class type (Clause 9) with a user-provided constructor (12.1), then the default constructor for T is called (and the initialization is ill-formed if T has no accessible default constructor);

— if T is a (possibly cv-qualified) non-union class type without a user-provided constructor, then the object is zero-initialized and, if T’s implicitly-declared default constructor is non-trivial, that constructor is called.

— if T is an array type, then each element is value-initialized; — otherwise, the object is zero-initialized.

For example:

#include <iostream>

class A {
    public:
    A(){}
    int i;
    int j;
};

class B {
    public:
    B() = default;
    int i;
    int j;
};

int main()
{
    for( int i = 0; i < 100; ++i) {
        A* pa = new A();
        B* pb = new B();
        std::cout << pa->i << "," << pa->j << std::endl;
        std::cout << pb->i << "," << pb->j << std::endl;
        delete pa;
        delete pb;
    }
  return 0;
}

possible result:

0,0
0,0
145084416,0
0,0
145084432,0
0,0
145084416,0
//...

http://ideone.com/k8mBrd