What is this crazy C++11 syntax ==> struct : bar {} foo {};?

Question

What could this possibly mean in C++11?

struct : bar {} foo {};

Answer 1

First, we'll take a bog-standard abstract UDT (User-Defined Type):

struct foo { virtual void f() = 0; }; // normal abstract type
foo obj;
// error: cannot declare variable 'obj' to be of abstract type 'foo'

Let's also recall that we can instantiate the UDT at the same time that we define it:

struct foo { foo() { cout << "!"; } };          // just a definition

struct foo { foo() { cout << "!"; } } instance; // so much more
// Output: "!"

Let's combine the examples, and recall that we can define a UDT that has no name:

struct { virtual void f() = 0; } instance; // unnamed abstract type
// error: cannot declare variable 'instance' to be of abstract type '<anonymous struct>'

We don't need the proof about the anonymous UDT any more, so we can lose the pure virtual function. Also renaming instance to foo, we're left with:

struct {} foo;

Getting close.

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Now, what if this anonymous UDT were to derive from some base?

struct bar {};       // base UDT
struct : bar {} foo; // anonymous derived UDT, and instance thereof

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Finally, C++11 introduces extended initialisers, such that we can do confusing things like this:

int x{0};

And this:

int x{};

And, finally, this:

struct : bar {} foo {};

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This is an unnamed struct deriving from bar, instantiated as foo with a blank initializer.

Answer 2

This defines:

• an anonymous struct,
• which is derived publicly from bar
• which (anonymously) defines nothing else but what it derived from bar
• and finally, an instance, called "foo" is created,
• with an empty initializer list

struct : bar {} foo {};