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Comment by reubenmorais | Hacker News Reader

This just declares a lambda with two parameters with default values. You can't do obj.x, you can't declare "objects" inside of obj, you can't return obj in the parent scope.

Why would you ever do:

  [x=1, y="foo"]{
    // use x, y
  }();

When you can just do:

  auto x = 1;
  auto y = "foo";
  // use x, y

?

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It is in the C++14 standard:

For every init-capture a non-static data member named by the identifier of the init-capture is declared in the closure type. This member is not a bit-field and not mutable. http://isocpp.org/files/papers/N3648.html

Actually the member can still be declared `private:` and the `obj.a.name` would not be portable. It appears GCC's implementation to make these members `public:` is conforming to c++11 but do not rely on this behavior.

I've been told in ##C++ that the current C++14 draft says: `An init-capture behaves as if it declares and explicitly captures a variable of the form “auto init-capture ;” whose declarative region is the lambda-expression’s compound-statement`

This change would make the named member access non-conforming

I have no idea if it is in the standard, but this actually compile with gcc 4.9 -std=c++1y

auto obj = [ x=1, y="foo" ]{}; obj.x; obj.y;

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