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Home/ Questions/Q 6324665
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Editorial Team
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Editorial Team
Asked: May 24, 20262026-05-24T16:44:52+00:00 2026-05-24T16:44:52+00:00

What could this possibly mean in C++11? struct : bar {} foo {};

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What could this possibly mean in C++11?

struct : bar {} foo {};
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  1. Editorial Team
    Editorial Team
    2026-05-24T16:44:53+00:00Added an answer on May 24, 2026 at 4:44 pm

    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.


    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
    

    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 {};
    

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

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