What does this C++ code mean?

Question

What does the following C++ code mean?

unsigned char a : 1; 
unsigned char b : 7;

I guess it creates two char a and b, and both of them should be one byte long, but I have no idea what the ": 1" and ": 7" part does.

Answer 1

The 1 and the 7 are bit sizes to limit the range of the values. They're typically found in structures and unions:

struct x {
    unsigned char a:1;
    unsigned char b:7;
} y;

creates an 8-bit value, one bit for a and 7 bits for b.

Typically used in C to access "compressed" values such as a 4-bit nybble which might be contained in the top half of an 8-bit char.

Update: For the language lawyers among us, the 9.6 section of the standard (C++0x, draft n2914) explains this in detail.

9.6 Bit-fields [class.bit]

1. A member-declarator of the form
        identifier_opt attribute-specifier_opt : constant-expression
   specifies a bit-field; its length is set off from the bit-field name by a colon. The optional attribute-specifier appertains to the entity being declared. The bit-field attribute is not part of the type of the class member. The constant-expression shall be an integral constant expression with a value greater than or equal to zero. The value of the integral constant expression may be larger than the number of bits in the object representation (3.9) of the bit-field’s type; in such cases the extra bits are used as padding bits and do not participate in the value representation (3.9) of the bit-field. Allocation of bit-fields within a class object is implementation-defined. Alignment of bit-fields is implementation-defined. Bit-fields are packed into some addressable allocation unit. [ Note: bit-fields straddle allocation units on some machines and not on others. Bit-fields are assigned right-to-left on some machines, left-to-right on others. — end note ]

2. A declaration for a bit-field that omits the identifier declares an unnamed bit-field. Unnamed bit-fields are not members and cannot be initialized. [ Note: an unnamed bit-field is useful for padding to conform to externally-imposed layouts. — end note ] As a special case, an unnamed bit-field with a width of zero specifies alignment of the next bit-field at an allocation unit boundary. Only when declaring an unnamed bit-field may the value of the constant-expression be equal to zero.

3. A bit-field shall not be a static member. A bit-field shall have integral or enumeration type (3.9.1). It is implementation-defined whether a plain (neither explicitly signed nor unsigned) char, short, int or long bit-field is signed or unsigned. A bool value can successfully be stored in a bit-field of any nonzero size. The address-of operator & shall not be applied to a bit-field, so there are no pointers to bit-fields. A non-const reference shall not be bound to a bit-field (8.5.3). [ Note: if the initializer for a reference of type const T& is an lvalue that refers to a bit-field, the reference is bound to a temporary initialized to hold the value of the bit-field; the reference is not bound to the bit-field directly. See 8.5.3. — end note ]

4. If the value true or false is stored into a bit-field of type bool of any size (including a one bit bit-field), the original bool value and the value of the bit-field shall compare equal. If the value of an enumerator is stored into a bit-field of the same enumeration type and the number of bits in the bit-field is large enough to hold all the values of that enumeration type (7.2), the original enumerator value and the value of the bit-field shall compare equal. [ Example:
        enum BOOL { FALSE=0, TRUE=1 };
     struct A {
          BOOL b:1;
     };
     A a;
     void f() {
          a.b = TRUE;
          if (a.b == TRUE) // yields true
          { /* ... */ }
     }
   - end example ]

Answer 2

I believe those would be bitfields.