25 / 2 = 12r1 (12 with a remainder of 1) 12 / 2 = 6r0 (6 with a remainder of 0) 6 / 2 = 3r0 (3 with a remainder of 0) 3 / 2 = 1r1 (1 with a remainder of 0) 1 / 2 = 0r1 (0...

Find a utility class to do this for you. Surely someone out there has written a BCD conversion utility for Java. Here you go. I Googled "BCD Java" and got this as the first result. Copying code here for future reference. public class BCD { /* * long number to...

Binary Coded Decimal is where a 4 bits are used to encode one digit of a number: 0000 = 0 0001 = 1 0010 = 2 0011 = 3 0100 = 4 0101 = 5 0110 = 6 0111 = 7 1000 = 8 1001 = 9 To encode a...

Ok, i found the solution, i simple changes this line: values3[i] = Byte.parseByte(values2[i], 2); //HERE IS THE PROBLEM For this one and all work perfect now! values3[i] = (byte) Integer.parseInt(values2[i], 2); If anyone wants to explain this i will be grateful. Regards!...

A generalized version: Suppose your number is stored in variable x. final int x = 345421; final String serialized = String.valueOf(x); final int[] array = new int[serialized.length()]; for (int i = 0; i < serialized.length(); i++) { array[i] = Integer.valueOf(String.valueOf(serialized.charAt(i))); } ...

Assuming your input is always between 0 and 99, inclusive: unsigned char hex = ((dec / 10) << 4) | (dec % 10); Simply take the upper digit and shift it left by one nibble, and or the lower digit in place....

In VHDL, a variable retains it's value between process re-entry. Thus, when you enter the process for the number X"000003", all your variables still have the value they add at the end of the processing of X"003565". A quick test shows that setting place to 1 at the start of...

There is a 1 byte datatype in (almost) every programming language. It is the character. It is actually the definition of a byte, that it can hold a default character. There is also a 1-byte (strictly speaking 1-octet) integer type in Fortran, accessible as integer(int8) where int8 is a constant...

You need to work with the two nybbles, multiplying the more significant nybble by ten and adding the less significant: uint8_t hex = 0x11; assert(((hex & 0xF0) >> 4) < 10); // More significant nybble is valid assert((hex & 0x0F) < 10); // Less significant nybble is valid int dec...

computer-science,computer-architecture,digit,bcd,vlsi

A Mod 4 in the original equation gives the two least significant bits of the BCD digit A. So, let's consider the two most and least significant bits of A separately, like so: X = A AND 3 = A Mod 4 (two least significant bits) Y = A AND...