Question

I’ve just done a test with bitfields, and the results are surprising me.

class test1 { public:     bool test_a:1;     bool test_b:1;     bool test_c:1;     bool test_d:1;     bool test_e:1;     bool test_f:1;     bool test_g:1;     bool test_h:1; };  class test2 { public:     int test_a:1;     int test_b:1;     int test_c:1;     int test_d:1;     int test_e:1;     int test_f:1;     int test_g:1;     int test_h:1; };  class test3 { public:     int test_a:1;     bool test_b:1;     int test_c:1;     bool test_d:1;     int test_e:1;     bool test_f:1;     int test_g:1;     bool test_h:1; }; 

The results were:-

sizeof(test1) = 1   // This is what I'd expect. 8 bits in a byte sizeof(test2) = 4   // Reasonable. Maybe padded out to the size of an int. sizeof(test3) = 16  // What??? 

Is this what you’d expect, or a compiler bug? (Codegear C++ Builder 2007, btw…)

Answer 1

your compiler has arranged all of the members of test3 on integer size boundaries. Once a block has been used for a given type (integer bit-field, or boolean bit-field), the compiler does not allocate any further bit fields of a different type until the next boundary.

I doubt it is a bug. It probably has something to do with the underlying architecture of your system.

edit:

c++ compilers will allocate bit-fields in memory as follows: several consecutive bit-field members of the same type will be allocated sequentially. As soon as a new type needs to be allocated, it will be aligned with the beginning of the next logical memory block. The next logical block will depend on your processor. Some processors can align to 8-bit boundaries, while others can only align to 16-bit boundaries.

In your test3, each member is of a different type than the one before it, so the memory allocation will be 8 * (the minimum logical block size on your system). In your case, the minimum block size is two bytes (16-bit), so the size of test3 is 8*2 = 16.

On a system that can allocate 8-bit blocks, I would expect the size to be 8.