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 numbers (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. 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.

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.

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 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.

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) Hide Solution

Binary digit	0	1	0	0	1	1	0	1
* Digit value	128	64	32	16	8	4	2	1
= Total (77)	0	64	0	0	8	4	0	1

The answer is 77.

2) Hide Solution

Using method 1:
93 / 2 = 46 r1
46 / 2 = 23 r0
23 / 2 = 11 r1
11 / 2 = 5 r1
5 / 2 = 2 r1
2 / 2 = 1 r0
1 / 2 = 0 r1

Working backwards from the remainders, 101 1101

Using method 2:
The largest power of 2 less than 93 is 64.

Is 93 >= 64? Yes, so the 64 bit is 1. 93 - 64 = 29.
Is 29 >= 32? No, so the 32 bit is 0.
Is 29 >= 16? Yes, so the 16 bit is 1. 29 - 16 = 13.
Is 13 >= 8? Yes, so the 8 bit is 1. 13 - 8 = 5.
Is 5 >= 4? Yes, so the 4 bit is 1. 5 - 4 = 1.
Is 1 >= 2? No, so the 2 bit is 0.
Is 1 >= 1? Yes, so the 1 bit is 1.

The answer is 0101 1101.

3) Hide Solution

We already know that 93 is 0101 1101 from the previous example.
For two’s complement, we invert the bits: 1010 0010
And add 1: 1010 0011

4) Hide Solution

Working right to left:
1010 0010 = (0 * 1) + (1 * 2) + (0 * 4) + (0 * 8) + (0 * 16) + (1 * 32) + (0 * 64) + (1 * 128) = 2 + 32 + 128 = 162.

The answer is 162.

5) Hide Solution

Since we’re told this number is in two’s complement, we can “undo” the two’s complement by inverting the bits and adding 1.
First, start with out binary number: 1010 0010
Flip the bits: 0101 1101
Add 1: 0101 1110

Convert to decimal: 64 + 16 + 8 + 4 + 2 = 94
Remember that this is a two’s complement #, and the original left bit was negative: -94

The answer is -94

6) Hide Solution

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#include <iostream>   // x is our number to test // pow is a power of 2 (e.g. 128, 64, 32, etc...) int printandDecrementBit(int x, int pow) {     // Test whether our x is greater than some power of 2 and print the bit     if (x >= pow)         std::cout << "1";     else         std::cout << "0";       // If x is greater than our power of 2, subtract the power of 2     if (x >= pow)         return x - pow;     else         return x; }   int main() {     std::cout << "Enter an integer between 0 and 255: ";     int x;     std::cin >> x;       x = printandDecrementBit(x, 128);     x = printandDecrementBit(x, 64);     x = printandDecrementBit(x, 32);     x = printandDecrementBit(x, 16);       std::cout << " ";       x = printandDecrementBit(x, 8);     x = printandDecrementBit(x, 4);     x = printandDecrementBit(x, 2);     x = printandDecrementBit(x, 1);       return 0; }