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3.7 — Converting between binary and decimal

In order to understand the bit manipulation operators, it is first necessary to understand how integers are represented in binary. We talked a little bit about this in section 2.4 -- Integers, and will expand upon it here.

Consider a normal decimal number, such as 5623. We intuitively understand that these digits mean (5 * 1000) + (6 * 100) + (2 * 10) + (3 * 1). Because there are 10 decimal numbers, the value of each digit increases by a factor of 10.

Binary numbers work the same way, except because there are only 2 binary digits (0 and 1), the value of each digit increases by a factor of 2. Just like commas are often used to make a large decimal number easy to read (e.g. 1,427,435), we often write binary numbers in groups of 4 bits to make them easier to read (e.g. 1101 0101).

As a reminder, in binary, we count from 0 to 15 like this:

Decimal Value Binary Value
0 0
1 1
2 10
3 11
4 100
5 101
6 110
7 111
8 1000
9 1001
10 1010
11 1011
12 1100
13 1101
14 1110
15 1111

Converting binary to decimal

In the following examples, we assume that we’re dealing with unsigned integers.

Consider the 8 bit (1 byte) binary number 0101 1110. 0101 1110 means (0 * 128) + (1 * 64) + (0 * 32) + (1 * 16) + (1 * 8) + (1 * 4) + (1 * 2) + (0 * 1). If we sum up all of these parts, we get the decimal number 64 + 16 + 8 + 4 + 2 = 94.

Here is the same process in table format. We multiply each binary digit by its digit value (determined by its position). Summing up all these values gives us the total.

Converting 0101 1110 to decimal:

Binary digit 0   1   0   1   1   1   1   0  
* Digit value

128 64 32 16 8 4 2 1
= Total (94)

0 64 0 16 8 4 2 0

Let’s convert 1001 0111 to decimal:

Binary digit 1   0   0   1   0   1   1   1  
* Digit value

128 64 32 16 8 4 2 1
= Total (151)

128 0 0 16 0 4 2 1

1001 0111 binary = 151 in decimal.

This can easily be extended to 16 or 32 bit binary numbers simply by adding more columns. Note that it’s easiest to start on the right end, and work your way left, multiplying the digit value by 2 as you go.

Method 1 for converting decimal to binary

Converting from decimal to binary is a little more tricky, but still pretty straightforward. There are two good methods to do this.

The first method involves continually dividing by 2, and writing down the remainders. The binary number is constructed at the end from the remainders, from the bottom up.

Converting 148 from decimal to binary (using r to denote a remainder):

148 / 2 = 74 r0
74 / 2 = 37 r0
37 / 2 = 18 r1
18 / 2 = 9 r0
9 / 2 = 4 r1
4 / 2 = 2 r0
2 / 2 = 1 r0
1 / 2 = 0 r1

Writing all of the remainders from the bottom up: 1001 0100

148 decimal = 1001 0100 binary.

You can verify this answer by converting the binary back to decimal:

(1 * 128) + (0 * 64) + (0 * 32) + (1 * 16) + (0 * 8) + (1 * 4) + (0 * 2) + (0 * 1) = 148

Method 2 for converting decimal to binary

The second method involves working backwards to figure out what each of the bits must be. This method can be easier with small binary numbers.

Consider the decimal number 148 again. What’s the largest power of 2 that’s smaller than 148? 128, so we’ll start there.

Is 148 >= 128? Yes, so the 128 bit must be 1. 148 - 128 = 20, which means we need to find bits worth 20 more.
Is 20 >= 64? No, so the 64 bit must be 0.
Is 20 >= 32? No, so the 32 bit must be 0.
Is 20 >= 16? Yes, so the 16 bit must be 1. 20 - 16 = 4, which means we need to find bits worth 4 more.

Is 4 >= 8? No, so the 8 bit must be 0.
Is 4 >= 4? Yes, so the 4 bit must be 1. 4 - 4 = 0, which means all the rest of the bits must be 0.

148 = (1 * 128) + (0 * 64) + (0 * 32) + (1 * 16) + (0 * 8) + (1 * 4) + (0 * 2) + (0 * 1) = 1001 0100

In table format:

Binary number 1   0   0   1   0   1   0   0  
* Digit value

128 64 32 16 8 4 2 1
= Total (148)

128 0 0 16 0 4 0 0

Another example

Let’s convert 117 to binary using method 1:

117 / 2 = 58 r1
58 / 2 = 29 r0
29 / 2 = 14 r1
14 / 2 = 7 r0
7 / 2 = 3 r1
3 / 2 = 1 r1
1 / 2 = 0 r1

Constructing the number from the remainders from the bottom up, 117 = 111 0101 binary

And using method 2:

The largest power of 2 less than 117 is 64.

Is 117 >= 64? Yes, so the 64 bit must be 1. 117 - 64 = 53.
Is 53 >= 32? Yes, so the 32 bit must be 1. 53 - 32 = 21.
Is 21 >= 16? Yes, so the 16 bit must be 1. 21 - 16 = 5.

Is 5 >= 8? No, so the 8 bit must be 0.
Is 5 >= 4? Yes, so the 4 bit must be 1. 5 - 4 = 1.
Is 1 >= 2? No, so the 2 bit must be 0.
Is 1 >= 1? Yes, so the 1 bit must be 1.

117 decimal = 111 0101 binary.

Adding in binary

In some cases (we’ll see one in just a moment), it’s useful to be able to add two binary numbers. Adding binary numbers is surprisingly easy (maybe even easier than adding decimal numbers), although it may seem odd at first because you’re not used to it.

Consider two small binary numbers:
0110 (6 in decimal) +
0111 (7 in decimal)

Let’s add these. First, line them up, as we have above. Then, starting from the right and working left, we add each column of digits, just like we do in a decimal number. However, because a binary digit can only be a 0 or a 1, there are only 4 possibilities:

  • 0 + 0 = 0
  • 0 + 1 = 1
  • 1 + 0 = 1
  • 1 + 1 = 0, carry a 1 over to the next column

Let’s do the first column:

0110 (6 in decimal) +
0111 (7 in decimal)
----
   1

0 + 1 = 1. Easy.

Second column:

 1
0110 (6 in decimal) +
0111 (7 in decimal)
----
  01

1 + 1 = 0, with a carried one into the next column

Third column:

11
0110 (6 in decimal) +
0111 (7 in decimal)
----
 101

This one is a little trickier. Normally, 1 + 1 = 0, with a carried one into the next column. However, we already have a 1 carried from the previous column, so we need to add 1. Thus, we end up with a 1 in this column, with a 1 carried over to the next column

Last column:

11
0110 (6 in decimal) +
0111 (7 in decimal)
----
1101

0 + 0 = 0, but there’s a carried 1, so we add 1. 1101 = 13 in decimal.

Now, how do we add 1 to any given binary number (such as 1011 0011)? The same as above, only the bottom number is binary 1.

       1  (carry column)
1011 0011 (original binary number)
0000 0001 (1 in binary)
---------
1011 0100

Signed numbers and two’s complement

In the above examples, we’ve dealt solely with unsigned integers. In this section, we’ll take a look at how signed numbers (which can be negative) are dealt with.

Signed integers are typically stored using a method known as two’s complement. In two’s complement, the leftmost (most significant) bit is used as the sign bit. A 0 sign bit means the number is positive, and a 1 sign bit means the number is negative.

Positive signed numbers are stored just like positive unsigned numbers (with the sign bit set to 0).

Negative signed numbers are stored as the inverse of the positive number, plus 1.

Converting integers to binary two’s complement

For example, here’s how we convert -5 to binary two’s complement:

First we figure out the binary representation for 5: 0000 0101
Then we invert all of the bits: 1111 1010
Then we add 1: 1111 1011

Converting -76 to binary:

Positive 76 in binary: 0100 1100
Invert all the bits: 1011 0011
Add 1: 1011 0100

Why do we add 1? Consider the number 0. If a negative value was simply represented as the inverse of the positive number, 0 would have two representations: 0000 0000 (positive zero) and 1111 1111 (negative zero). By adding 1, 1111 1111 intentionally overflows and becomes 0000 0000. This prevents 0 from having two representations, and simplifies some of the internal logic needed to do arithmetic with negative numbers.

Converting binary two’s complement to integers

To convert a two’s complement binary number back into decimal, first look at the sign bit.

If the sign bit is 0, just convert the number as shown for unsigned numbers above.

If the sign bit is 1, then we invert the bits, add 1, then convert to decimal, then make that decimal number negative (because the sign bit was originally negative).

For example, to convert 1001 1110 from two’s complement into a decimal number:
Given: 1001 1110
Invert the bits: 0110 0001
Add 1: 0110 0010
Convert to decimal: (0 * 128) + (1 * 64) + (1 * 32) + (0 * 16) + (0 * 8) + (0 * 4) + (1 * 2) + (0 * 1) = 64 + 32 + 2 = 98
Since the original sign bit was negative, the final value is -98.

If adding in binary is difficult for you, you can convert to decimal first, and then add 1.

Why types matter

Consider the binary value 1011 0100. What value does this represent? You’d probably say 180, and if this were standard unsigned binary number, you’d be right.

However, if this value was stored using two’s complement, it would be -76.

And if the value were encoded some other way, it could be something else entirely.

So how does C++ know whether to print a variable containing binary 1011 0100 as 180 or -76?

Way back in section 2.1 -- Basic addressing and variable declaration, we said, “When you assign a value to a data type, the compiler and CPU takes care of the details of encoding your value into the appropriate sequence of bits for that data type. When you ask for your value back, your number is “reconstituted” from the sequence of bits in memory.”

So the answer is: it uses the type of the variable to convert the underlying binary representation back into the expected form. So if the variable type was an unsigned integer, it would know that 1011 0100 was standard binary, and should be printed as 180. If the variable was a signed integer, it would know that 1011 0100 was encoded using two’s complement (assuming that’s what it was using), and should be printed as -76.

What about converting floating point numbers from/to binary?

How floating point numbers get converted from/to binary is quite a bit more complicated, and not something you’re likely to ever need to know. However, if you’re curious, see this site, which does a good job of explaining the topic in detail.

Quiz

1) Convert 0100 1101 to decimal.
2) Convert 93 to an 8-bit unsigned binary number.
3) Convert -93 to an 8-bit signed binary number (using two’s complement).
4) Convert 1010 0010 to an unsigned decimal number.
5) Convert 1010 0010 to a signed decimal number (assume two’s complement).

6) Write a program that asks the user to input a number between 0 and 255. Print this number as an 8-bit binary number (of the form #### ####). Don’t use any bitwise operators.

Hint: Use method 2. Assume the largest power of 2 is 128.
Hint: Write a function to test whether your input number is greater than some power of 2. If so, print ‘1’ and return your number minus the power of 2.

Quiz answers

1) Show Solution

2) Show Solution

3) Show Solution

4) Show Solution

5) Show Solution

6) Show Solution

3.8 -- Bitwise operators
Index
3.6 -- Logical operators

346 comments to 3.7 — Converting between binary and decimal

  • Benur21

    "In the following examples, we assume that we’re dealing with unsigned integers."

    Integers with 1 byte?

    • Alex

      No, integers that are always positive. Revisit lesson 2.4.

      • Benur21

        But integers have 4 bytes (or 2 bytes minimum). I know unsigned doesn't have negative.

        • Alex

          It doesn't matter how big the integers are. The binary representation of a number doesn't change based on integer size (the high bits are just padded with 0s).

          For example, with a 4-bit integer, the number 3 is represented as: 0011.
          With an 8-bit integer, the number 3 is represented as 0000 0011.
          With a 16-bit integer, the number 3 is represented as 0000 0000 0000 0011.

  • Benur21

    "For example, to convert 1001 1110 from two’s complement into a decimal number:
    Given: 1001 1110
    Invert the bits: 0110 0001
    Add 1: 0110 0010
    Convert to decimal: (0 * 128) + (1 * 64) + (1 * 32) + (0 * 16) + (0 * 8) + (0 * 4) + (1 * 2) + (0 * 1) = 64 + 32 + 2 = 98
    Since the original sign bit was negative, the final value is -98."

    Where is the sign bit?

      • Benur21

        The one that was inverted? Isn't it part of the number? Then why is unsigned 1 byte 0-255 and signed 1 byte 0-127?

        • For unsigned numbers, the first bit is part of the number.
          For signed numbers, the first bit is the sign bit.
          Signed 1 byte is -128 to 127.

          • Benur21

            In this case, 1001 1110, 1 is the sign bit, and 001 110 is the number, right? Then the sign bit shouldn't be inverted and the (0 * 128) part should be removed, as it is not part of the number, it's the sign bit. Right? Because if it was still part of the number then we would have 0-255 instead of 0-127, which is wrong for a signed byte.

            • > In this case, 1001 1110, 1 is the sign bit, and 001 110 is the number, right?
              No. If the sign bit is set, you need to invert the entire number and add 1. Re-read the section about two's complement.

              > Then the sign bit shouldn't be inverted and the (0 * 128) part should be removed
              After inverting and adding 1, the first number could again be 1, but this time it's part of the number and no longer considered a sign.

              > 0-127
              There is no type with this range. It's either 0 to 255 or -128 to 127

              • Benur21

                "For example, to convert 1001 1110 from two’s complement into a decimal number:
                Given: 1001 1110
                Invert the bits: 0110 0001
                Add 1: 0110 0010
                Convert to decimal: (0 * 128) + (1 * 64) + (1 * 32) + (0 * 16) + (0 * 8) + (0 * 4) + (1 * 2) + (0 * 1) = 64 + 32 + 2 = 98
                Since the original sign bit was negative, the final value is -98."

                So when the byte is signed and the bit sign is 1, it will be able to be 0xxx xxxx, which will translate to a number between 0 and 127 (or 1 and 128 if you add 1). When bit sign is 0 (positive), it will be able to be 1xxx xxxx, which translates to between 128 and 255 (or 129 and 256 (0) if you add 1). So you can't have positive less than 128 (and bigger than 0)??

                • > when the byte is signed and the bit sign is 1, it will be able to be 0xxx xxxx
                  But you add 1, so it's either 0xxx xxxx or 1000 0000, but at least 0000 0001, which is 1 to 128, or -1 to -128, because the sign bit was negative. If the sign bit is 0, the range is 0000 0000 to 0111 1111 or 0 to 127.

                  > When bit sign is 0 (positive), it will be able to be 1xxx xxxx
                  You only invert if the sign bit is 1. If it's 0, the number stays as it is.

                  • Benur21

                    > You only invert if the sign bit is 1. If it's 0, the number stays as it is.

                    Well, then the "(0 * 128)" is optional, as it will always be zero.
                    Edit: Ahh, nevermind, I read what you said above about add 1.
                    Got it, thanks for helping.

  • Japlinho

    Here is my code:

  • Shawn

    This is how I did it.

  • Piotr

    I've wrote three functions: one for data input (unnecessary in this case), one for calculating the power of 2 and one for calculating and printing binary form and was so proud of myself. Then I've checked your solution and seen that it is so much more elegant and concise!

  • Jean Pi

    Hi!

    Sorry if this outside the scope of this site, but I was wondering why we encode negative numbers this way.
    If my understanding is correct, to write -5 in binary over a byte, we use 7 bits to write (-5 % 128) = 123, and add a 1 in front to indicate the number is negative. 0111 1011 is 123, so -5 is 1111 1011.
    Ultimately, if we interpret the 1 in front as we do for unsigned numebrs, it's as if we replaced 251 by -5. So the sequence of numbers goes like this: 0,1,2... 127,-128,-127,... -1.

    But doesn't this mean that -5 or any other negative number is not written the same way over 1 byte as it is over 2 bytes? Moreover, why do we use this particular sequence instead of another one? (for instance, 0,1,2...127,-1...-128, or even -128,-127,...127)

    Agin, sorry to bother you with this, I realize it has nothing to do with actual coding. Thanks for your efforts!

    • Hi Jean!

      > If my understanding is correct
      I don't know if your way of conversion is correct, here's how I learned it:

      The standard doesn't specify a specific negative number representation. Compiler developers are free to use other representations. The most common representation is two's complement (What I did above)

      > But doesn't this mean that -5 or any other negative number is not written the same way over 1 byte as it is over 2 bytes?
      That's what it means.

      > Moreover, why do we use this particular sequence
      The sequence is determined by the binary representation. You always have

      And convert that into decimal. Naturally, 0 is the first number. When using two's complement, the number following 127 (0111 1111) is -128 (1000 0000).

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