04-03. Denary to binary

This programme will convert a denary number into a binary number.

We'll introduce you to modulo and integer division.

As always, we start with the boilerplate code in the main programme.

We'll output a message asking the user to enter a denary number.

We'll store this in an integer variable called number and use convert to cast to an integer the line input from the keyboard.

Next, we're going to output the result.

So console, writeline, bracket, quote, followed by a message to explain what the output will be and the output from the function we'll write in a minute called then two bin, which is short for denary to binary.

This will require the number, so open bracket, the number we input.

Close bracket, close bracket.

Now we need to define the function dent that does the conversion static string, then to bin, which will have the parameter passed in number, which is an integer.

Worth noting here that we're not going to return an actual number, we're going to return a string which will have the characters for zeros and ones representing our binary output.

So the algorithm works like this.

We keep dividing the number by two and rounding down.

If the number was exactly divisible by 2, we want to join a 0 to the binary number and if not, we'll join a 1.

We do this while the number is above 0.

So a string binary that will store our answer is assigned to be an empty string.

To begin with, strings are qualified with quotes, so quote, quote, while number is greater than zero.

The remainder is the number divided by two, but the remainder, not decimal places.

Getting the remainder from a division is achieved using the modulo operator and in C, that's the percentage symbol.

Imagine you've got six suites and divide by two people.

Six divided by two is three sweets each, with no sweets left over, no remainder.

however, seven sweets divided by two is three, each with one left over.

The remainder, is one.

So six modulo two is zero, but seven modulo two is one.

Therefore, what we get as a result of dividing by 2 is either 0 or 1 as a remainder or modulus.

In our programme, remainder is 0 or 1.

At this point we want to join this to our binary string.

But because remainder is a number and binary is a string, we have to cast remainder to become a string to be able to join or concatenate it.

Remember, the binary code inside a computer for a number is different to the binary code for the characters 0 or 1.

Numbers are used for maths and strings are used for output to the screen.

Binary becomes the string of the remainder.

plus whatever the binary number is.

Now, interestingly, we can use the plus operator to both add numbers and concatenate strings together.

Number equals itself divided by two.

Now, notice a clever trick here.

Number was declared as an integer parameter when we defined the thentobin function.

This means if there are any decimal places after a division, they cannot be stored and are lost.

You might expect a syntax error here.

usually we have to cast one data type into another using convert to.

However, with simple division, this is not actually necessary.

Truncating a double number to an integer when dividing is also known as integer division.

Once the condition in the while loop is true, it will end, so we can return the string of characters we called binary that's representing our binary number.

This is one of those algorithms that illustrates why people find programming so difficult.

When we explained how the algorithm was going to work, we said if the remainder is zero, yet in our programme there is no if command.

That's because remainder can only be 0 or 1 after the modulo division, and we want to join it to the binary number either way.

So we didn't need to use an if we could.

But it's unnecessary because the joining of either 0 or 1 will happen.

Whatever the case, let's give the programme a run.

With the number 14, the binary is 11, 10 and.

And that's correct for unsigned binary.