Default values to bitfield elements

Question

In C++11 one can do

struct S {int i = 42;};

and if one forgets to initialize the member i it gets default initialized to 42. I Just tried this with bitfields as

struct S {int i = 42 : 5;};

and am getting

error: expected ';' before ':' token

Does this feature exist for bitfield members and if so, how do I use it?

Answer 1

The syntax for bit field initialization is

 struct S {int i: 5 = 42;};

and is only available in c++20: -std=c++2a for now

Answer 2

No, bit-fields do not allow an initializer as part of the member declaration. You can see this in the part of the grammar that describes class members (C++11 and later, [class.mem]):

member-declarator:
    declarator virt-specifier-seqopt pure-specifier_opt
    declarator brace-or-equal-initializer_opt
    identifieropt attribute-specifier-seq_opt : constant-expression

The third form is the grammar for a bit-field declaration, but only the second form lists the brace-or-equal-initializer.

Answer 3

You can also use a constructor to initialize a bitfield like this:

struct Foo {
    Foo () : i {15} {}

    int i : 5;
};

Foo foo;
cout << foo.i << endl; // 15

You can see it here