Question

I wonder if the following construction is possible:

typedef std::bitset<4> BIT4;
BIT4* x=new BIT4[3];
BIT4* y=new BIT4[5];

My concerns are:

(1) is the array of BIT4 a legal construction?

(2) if yes, how the pointer will address these arrays of size 3 and 5? last element of these arrays seems to take 1/2 byte, is another half just lost and we start the next array from a new byte or how does it work?

Thank you for your help!

Answer 1

What you are trying to do as I understand is

use 12 bits for 3 elements in an array called x

use 20 bits for 5 elements in an array called y

This is wrong as you will end up using:

• 3 bytes for x
• 5 bytes for y

if you want to achieve what you want you need to do it like this:

std::bitset<4*3> x;   // this will occupy 12 + 4 Padding Bits = 2 bytes
std::bitset<4*5> y;   // this will occupy 20 + 4 Padding Bits = 3 bytes

and you can of course allocate it dynamically if you want to.