Now eight shifts have been performed, so the algorithm terminates. If any digit is five or above, three is added to ensure the value 'carries' in base 10. Shifting left accomplishes both tasks simultaneously. The increment ensures that a value of 5, incremented and left-shifted, becomes 16 (10000), thus correctly 'carrying' into the next BCD digit.Įssentially, the algorithm operates by doubling the BCD value on the left each iteration and adding either one or zero according to the original bit pattern. On each iteration, any BCD digit which is at least 5 (0101 in binary) is incremented by 3 (0011) then the entire scratch space is left-shifted one bit. The scratch space is initialized to all zeros, and then the value to be converted is copied into the 'original register' space on the right. The diagram above shows the binary representation of 243 10 in the original register, and the BCD representation of 243 on the left. For example, if the original number to be converted is eight bits wide, the scratch space would be partitioned as follows: Then partition the scratch space into BCD digits (on the left) and the original register (on the right). It takes a maximum of 4 bits in binary to store each decimal digit. Reserve a scratch space wide enough to hold both the original number and its BCD representation n + 4× ceil( n/3) bits will be enough. Suppose the original number to be converted is stored in a register that is n bits wide. It is also known as the shift-and-add-3 algorithm, and can be implemented using a small number of gates in computer hardware, but at the expense of high latency. In computer science, the double dabble algorithm is used to convert binary numbers into binary-coded decimal (BCD) notation.

Algorithm to convert binary numbers to BCD

Verilog convert binary to decimal